$(A,t)\in{\overline{C}}$, the space of outgoing directions is
${\mathbb{P}}{\rm Ext}^{1}(A^{\vee}(\boldsymbol{\mathsf{p}}),A)$ which is by
definition the fiber of $D$ over $(A,t)$. As we saw in Lemma 3.9(b) and we
will see again by a different argument in Proposition 4.5 below,
$D\cong{\overline{C}}\times{\mathbb{P}}^{1}$. Each ${\mathbb{P}}^{1}$ fiber of
$D\to{\overline{C}}$ maps to the space of extensions which is the general form
of a line in $X=X_{1}$. Both the lines and the horizontal ${\overline{C}}$’s
map to positive degree curves, and an effective divisor on the product $D$ is
a positive sum of vertical and horizontal ones, so any effective divisor on
$D$ maps to a positive degree curve in $X$, in particular it is not
contracted. For generic injectivity, we note that a general wobbly bundle $E$
has a one dimensional space of nilpotent Higgs fields and if we $\theta$ is
any non-zero nilpotent Higgs field, we get a well defined line subbundle
$A=\ker\theta E$ and a realization of $E$ as an extension of
$A^{\vee}(\boldsymbol{\mathsf{p}})$ by $A$. This determines a unque point in
$D$ which shows that the map $D\to{\rm Wob}$ has a rational section. Since $D$
is irreducible this implies that $D\to{\rm Wob}$ is a birational morphism.
Alternatively, we can see the generic injectivity from the geometry of the
quadric line complex elucidated in section 2.5. Recall that at a general point
$w$ of ${\rm Wob}$ the four lines in $X$ through that point include one with
multiplicity $2$. That line corresponds to the point
$A\in{\overline{C}}\subset\operatorname{Jac}^{0}(C)$, and the position of $w$
on it corresponds to the extension class, so we can recover the data of a
point of $D$. It follows that $D$ is the normalization of ${\rm Wob}$. ∎

######  Lemma 4.4.

The divisor class of the wobbly locus is $[{\rm Wob}]=8H$.

###### Proof.

We will consider lines $\ell\subset X$ and show that $\ell$ intersects $W$ in
$8$ points. In the moduli point of view on $X$ a line corresponds to fixing a
line bundle $L$ of degree $0$ and looking at the set of bundles $E$ fitting
into an extension

$0\rightarrow L\rightarrow E\rightarrow
L^{\vee}(\boldsymbol{\mathsf{p}})\rightarrow 0.$

Such an extension is wobbly if there is $A\rightarrow E$ and $t\in C$ with
$A^{\otimes 2}={\mathcal{O}}(t-\boldsymbol{\mathsf{p}})$. Assuming that $L$
does not satisfy this condition, it means that
$A=L^{\vee}(\boldsymbol{\mathsf{p}}-y)$, so $L^{\otimes
2}(2y-2\boldsymbol{\mathsf{p}})={\mathcal{O}}(\boldsymbol{\mathsf{p}}-t)$ in
other words $L^{\otimes 2}={\mathcal{O}}(3\boldsymbol{\mathsf{p}}-t-2y)$, and
the pullback of the extension to $A$ splits. For a choice of $(y,t)$ there
will be a unique extension up to scalars, i.e. a unique point on the line,
such that the pullback extension splits, so we need to look at the number of
solutions of the equation
$L^{\otimes-2}(3\boldsymbol{\mathsf{p}})={\mathcal{O}}(2y+t)$. This is the
number of branch points of a trigonal map $C\rightarrow{\mathbb{P}}^{1}$,
which is $8$. Thus $\ell\cap{\rm Wob}$ has $8$ points for a general line
$\ell$. We will discuss the trigonal covers more in subsection 4.5 below. ∎

The basic structure of ${\rm Wob}$ is captured by the following proposition.

######  Proposition 4.5.

We have $D\cong{\overline{C}}\times{\mathbb{P}}^{1}$ and the factor
${\mathbb{P}}^{1}$ is naturally identified with the base of the hyperelliptic
covering ${\mathbb{P}}^{1}=C/\boldsymbol{\iota}_{C}$. The map
$D\rightarrow{\rm Wob}$ identifies together pairs of points on $6$ curves of
the form ${\overline{C}}\times\\{\boldsymbol{\mathsf{x}}_{i}\\}$, yielding a
nodal locus in ${\rm Wob}$. Furthermore, on the graph of the natural
projection ${\overline{C}}\rightarrow{\mathbb{P}}^{1}$ the map $D\rightarrow
X$ is non-immersive, yielding a cuspidal locus in ${\rm Wob}$.

###### Proof.

Recall that
$\boldsymbol{\mathsf{x}}_{1},\ldots,\boldsymbol{\mathsf{x}}_{6}\in{\mathbb{P}}^{1}$
denoted the branch points for the hyperelliptic map
$\mathsf{h}_{C}:C\to{\mathbb{P}}^{1}$, i.e. the images of the Weierstrass
points $\boldsymbol{\mathsf{p}}_{1},\ldots,\boldsymbol{\mathsf{p}}_{6}\in C$;
also we chose $\boldsymbol{\mathsf{p}}:=\boldsymbol{\mathsf{p}}_{1}$ to fix
the determinant $\det(E)\cong{\mathcal{O}}_{C}(\boldsymbol{\mathsf{p}})$ of
our stable vector bundles.

The statement that the ${\mathbb{P}}^{1}$-fiber of $D$ is naturally identified
with the hyperelliptic ${\mathbb{P}}^{1}$ was already checked in Lemma 3.9(b)
but we recall the argument here for convenience. A point in ${\overline{C}}$
is a line bundle $L$ such that $L^{\otimes 2}(\boldsymbol{\mathsf{p}})$ is
effective, or equivalently such that $L^{\otimes
2}\cong{\mathcal{O}}_{C}(t-\boldsymbol{\mathsf{p}})$ for some point $t\in C$.
Thus $L$ defines uniquely the point $t$.

A point in $D$ is a pair $(L,\eta)$ where $L\in{\overline{C}}$ and
$\eta\in{\mathbb{P}}\operatorname{Ext}^{1}(L^{\vee}(\boldsymbol{\mathsf{p}}),L)$,
defining the extension

$0\rightarrow L\rightarrow E\rightarrow
L^{\vee}(\boldsymbol{\mathsf{p}})\rightarrow 0.$

We note that

$\operatorname{Ext}^{1}(L^{\vee}(\boldsymbol{\mathsf{p}}),L)=H^{1}(L^{\otimes
2}(-\boldsymbol{\mathsf{p}}))\cong
H^{1}({\mathcal{O}}_{C}(t-2\boldsymbol{\mathsf{p}}))\cong
H^{1}({\mathcal{O}}_{C}(-t^{\prime}))$

where $t^{\prime}$ is the image of $t$ under the hyperelliptic involution. The
map ${\mathcal{O}}_{C}(-t^{\prime})\rightarrow{\mathcal{O}}_{C}$ induces an
isomorphism on $H^{1}$ so we can write

$\operatorname{Ext}^{1}(L^{\vee}(\boldsymbol{\mathsf{p}}),L)\cong
H^{1}({\mathcal{O}}_{C}).$

The scalar multiplying this isomorphism is not canonical, it depends on
something and indeed that leads (as we have discussed) to the degree $128$ of
the normal bundle of ${\widehat{C}}$ in ${\mathcal{P}}$. However, it does give
a canonical isomorphism

${\mathbb{P}}\operatorname{Ext}^{1}(L^{\vee}(\boldsymbol{\mathsf{p}}),L)\cong{\mathbb{P}}H^{1}({\mathcal{O}}_{C})={\mathbb{P}}^{1}.$

We note that if $y\in C$ then the map

$H^{1}({\mathcal{O}}_{C})\rightarrow H^{1}({\mathcal{O}}_{C}(y))$

is a quotient of rank $1$ corresponding to the inage
$\mathsf{h}_{C}(y)\in{\mathbb{P}}^{1}$ of $y$. This is the identification
between the ${\mathbb{P}}^{1}$ factor of $D$, and the base of the
hyperelliptic projection of $C$.

We would like to understand when two points of $D$ correspond to the same
bundle. Suppose given $(L,\eta)\in D$, corresponding to an extension as above.
We look for a new line subbundle $L_{1}\subset E$ such that
$L_{1}\in{\overline{C}}$. We note first of all that if $L\neq L_{1}$ as
subbundles in $E$, then there is an inclusion

$L_{1}\hookrightarrow L^{\vee}(\boldsymbol{\mathsf{p}}),$

so we may write

$L_{1}=L^{\vee}(\boldsymbol{\mathsf{p}}-y).$

We have

$L_{1}^{\otimes
2}=L^{\otimes-2}(2\boldsymbol{\mathsf{p}}-2y)={\mathcal{O}}_{C}(\boldsymbol{\mathsf{p}}-t)(2\boldsymbol{\mathsf{p}}-2y)={\mathcal{O}}_{C}(3\boldsymbol{\mathsf{p}}-t-2y)$

so the condition $L_{1}\in{\overline{C}}$ becomes

${\mathcal{O}}_{C}(3\boldsymbol{\mathsf{p}}-t-2y)\cong{\mathcal{O}}_{C}(t_{1}-\boldsymbol{\mathsf{p}})$

for some point $t_{1}\in C$. In other words,

$4\boldsymbol{\mathsf{p}}\sim t+t_{1}+2y.$

One may notice that for our hyperelliptic curve, any time that $a+b+c+d\sim
2\omega_{C}=4\boldsymbol{\mathsf{p}}$, we have
$a+b+c+d=x+x^{\prime}+y+y^{\prime}$ such that $x,x^{\prime}$ (respectively
$y,y^{\prime}$) are paired under the hyperelliptic involution
$\boldsymbol{\iota}_{C}$.

Given the form $t+t_{1}+2y$, there are only two possibilities:

  * (i)

either $t=t_{1}$, and $t$ is $\boldsymbol{\iota}_{C}$-paired with $y$,

  * (ii)

or $(t,t_{1})$ are $\boldsymbol{\iota}_{C}$-paired, and
$y=\boldsymbol{\mathsf{p}}_{i}$ is a Weierstrass point on $C$.

Case (i): If we are in the first case we get

$L_{1}=L^{\vee}(\boldsymbol{\mathsf{p}}-y)=L^{\vee}(\boldsymbol{\mathsf{p}}-(2\boldsymbol{\mathsf{p}}-t))=L^{\vee}(t-\boldsymbol{\mathsf{p}})$

and so we must have $L_{1}\cong L$, that is $L$ and $L_{1}$ correspond to two
different embedings of the same line bundle in $E$. However, for $L_{1}\cong
L$ the space $\operatorname{Ext}^{1}(L_{1},L)$ of extensions of $L_{1}$ by $L$
has dimension $2$ (it is $H^{1}({\mathcal{O}}_{C})$. Furthermore, the map on
extension sapces

$\operatorname{Ext}^{1}(L^{\vee}(\boldsymbol{\mathsf{p}}),L)\to\operatorname{Ext}^{1}(L_{1},L)$

induced by the inclusion $L_{1}\hookrightarrow
L^{\vee}(\boldsymbol{\mathsf{p}})$, is identified with the map on cohomology

$H^{1}(L^{\otimes
2}(-\boldsymbol{\mathsf{p}}))=H^{1}(\mathcal{O}_{C}(t-2\boldsymbol{\mathsf{p}}))=H^{1}(\mathcal{O}_{C}(-t^{\prime}))\to
H^{1}(\mathcal{O}_{C})$

induced by the inclusion
$\mathcal{O}_{C}(-t^{\prime})\hookrightarrow\mathcal{O}_{C}$, which is an
isomorphism. Hence if $L_{1}\cong L$, then no choice of bundle $E$ admits a
lifting from $L_{1}$ into $E$. Therefore, case (i) does not lead to any
identification.

As we will see below, it will however lead to an identification
infinitesimally, yielding the cuspidal locus. We ignore case (i) for now.

Case (ii): In this case, we have $y=p_{i}$ and $t_{1}=t^{\prime}$. This gives
a linear equivalence $2\boldsymbol{\mathsf{p}}\sim 2y$ and we get

$L_{1}=L^{\vee}(\boldsymbol{\mathsf{p}}-p_{i})=L^{\vee}\otimes\mathfrak{a}_{i}$

where
$\mathfrak{a}_{i}={\mathcal{O}}_{C}(\boldsymbol{\mathsf{p}}-\boldsymbol{\mathsf{p}}_{i})$
is a $2$-torsion point. There are six possibilities in this case. Let us look
at what is the extension $\eta$ such that $L_{1}$ lifts into $E$. The
extension $\eta$ should be in the kernel of the map

$\operatorname{Ext}^{1}(L^{\vee}(\boldsymbol{\mathsf{p}}),L)\rightarrow\operatorname{Ext}^{1}(L_{1},L)$

i.e. the map

$H^{1}({\mathcal{O}}(-t^{\prime}))\rightarrow
H^{1}({\mathcal{O}}(-t^{\prime}+\boldsymbol{\mathsf{p}}_{i})).$

We note that this map is isomorphic to the map

$H^{1}({\mathcal{O}}_{C})\rightarrow
H^{1}({\mathcal{O}}_{C}(\boldsymbol{\mathsf{p}}_{i}))$

so it means that $\eta$ should be the extension corresponding to the branch
point $\boldsymbol{\mathsf{x}}_{i}\in{\mathbb{P}}^{1}$.

We obtain $6$ glueings, one for each point $\boldsymbol{\mathsf{x}}_{i}$, with
the point $(L,\boldsymbol{\mathsf{x}}_{i})$ being glued to
$(L\otimes\mathfrak{a},\boldsymbol{\mathsf{x}}_{i})$. In other words the
surface $D$ is glued to itself along each of the $6$ curves
${\overline{C}}\times\\{\boldsymbol{\mathsf{x}}_{i}\\}$, $i=1,\ldots,6$. On
each curve ${\overline{C}}\times\\{\boldsymbol{\mathsf{x}}_{i}\\}$, the
glueing map is the automorphism of ${\overline{C}}$ given by tensoring with
the $2$-torsion point $\mathfrak{a}_{i}$. This is the first part of the
statement in the proposition.

The analysis of the cuspidal locus, to conclude the proof of the proposition,
will be done later in subsection 4.4. ∎

###  4.2 Computations in degree one

By convention all answers to Chern calculations are truncated to the terms of
degree $\leq 2$.

By Lemma 3.2) and Theorem 3.6 the curve ${\widehat{C}}$ embeds in
${\mathcal{P}}$ and the blow-up
$\boldsymbol{\varepsilon}:Y\rightarrow{\mathcal{P}}$ of ${\mathcal{P}}$ along
${\widehat{C}}$ resolves the natural rational map
$\pi_{*}(-):{\mathcal{P}}\dashrightarrow X$ producing a finite morphism
$f:Y\to X$. Let $\boldsymbol{\mathit{E}}\subset Y$ denote the exceptional
locus. It is a ${\mathbb{P}}^{1}$-bundle over ${\widehat{C}}$, namely
$\boldsymbol{\mathit{E}}\cong{\mathbb{P}}(N)$ (this being the projective
bundle of subspaces), where $N:=N_{{\widehat{C}}/{\mathcal{P}}}$. In fact by
Lemma 3.9(b) we have
$N=H^{1}(\mathcal{O}_{C})\otimes\widehat{{\mathsf{sq}}}^{*}\pi^{*}\omega_{C}$
and so $\boldsymbol{\mathit{E}}={\widehat{C}}\times{\mathbb{P}}^{1}$ where the
second factor is the hyperelliptic ${\mathbb{P}}^{1}$ for $C$.

######  Remark 4.6.

The images in $X$ of the ${\mathbb{P}}^{1}$’s that are fibers of
$b_{\boldsymbol{\mathit{E}}}:\boldsymbol{\mathit{E}}\rightarrow{\widehat{C}}$
are lines in $X$. Indeed, these fibers are isomorphic to the projectivized
bundle of outgoing directions along ${\overline{C}}$ which map to lines in
${\rm Wob}$, see Lemma 4.3.

Let $H$ denote the hyperplane class on $X$. Since $X\subset{\mathbb{P}}^{5}$
is a degree $4$ subvariety of dimension $3$, we get $H^{3}=4$.

Let $\boldsymbol{\mathit{F}}=f^{*}H$ be the inverse image of the class $H$ on
$Y$. We know that for a generic Hitchin fiber the Neron-Severi group of
${\mathcal{P}}$ is ${\mathbb{Z}}$, so the Neron-Severi group of $Y$ is
generated rationally by the classes $\boldsymbol{\mathit{E}}$ and
$\boldsymbol{\mathit{F}}$.

######  Proposition 4.7.

The Todd classes for the degree $1$ moduli spaces are:

${\rm td}(TX)=1+H+7H^{2}/12,$ $\pi^{*}{\rm
td}(TX)^{-1}=1-\boldsymbol{\mathit{F}}+5\boldsymbol{\mathit{F}}^{2}/12,$ ${\rm
td}(TY)=1-\boldsymbol{\mathit{E}}/2+(\boldsymbol{\mathit{E}}^{2}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}})/9,$
${\rm
td}(Y/X)=(1-\boldsymbol{\mathit{F}}+5\boldsymbol{\mathit{F}}^{2}/12)(1-\boldsymbol{\mathit{E}}/2+(\boldsymbol{\mathit{E}}^{2}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}})/9).$

###### Proof.

Recall from Section 3.12, the Chern class calculations for the intersection of
two quadrics $X$ say

$c_{1}(TX)=2H,\;\;\;\;c_{2}(TX)=3H^{2},$

and

${\rm td}(TX)=1+c_{1}/2+(c_{1}^{2}+c_{2})/12$ $=1+H+7H^{2}/12.$

A computation shows that

$(1+H+7H^{2}/12)(1-H+5H^{2}/12)=1,$

thus

${\rm td}(TX)^{-1}=1-H+5H^{2}/12$

which pulls back to the same formula with $\boldsymbol{\mathit{F}}$ on $Y$.

For the blow-up $Y$ of an abelian threefold ${\mathcal{P}}$ along the curve
${\widehat{C}}$ we need to know the degree of the normal bundle. This degree
is $2g_{{\widehat{C}}}-2$ for the embedding of a curve in an abelian variety,
thus by either the explicit description of $N$ in Lemma 3.9(b) or by the genus
calculation of Lemma 4.1 we have

$\delta={\rm deg}(N)=128.$

This could also be seen from the intersection number calculations above, one
can see that it is the same as $-\boldsymbol{\mathit{E}}^{3}$.

Again from the calculations of Section 3.12 we have

$c_{1}(TY)=-\boldsymbol{\mathit{E}},$
$c_{2}(TY)=\delta\mathsf{fib}-\boldsymbol{\mathit{E}}^{2},$

and

${\rm td}(TY)=1+c_{1}/2+(c_{1}^{2}+c_{2})/12$
$=1-\boldsymbol{\mathit{E}}/2+\delta\mathsf{fib}/12.$

We note that
$(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})|_{\boldsymbol{\mathit{E}}}$
is a divisor on $\boldsymbol{\mathit{E}}$ whose intersection with a fiber is
zero, indeed a fiber intersect $\boldsymbol{\mathit{E}}$ is $-1$ and it
intersects $\boldsymbol{\mathit{F}}$ in $1$ point since the image of a fiber
is a line contained in $X$ (Remark 4.6). Therefore
$(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})\cdot\boldsymbol{\mathit{E}}$
is a sum of fibers. The number may be calculated as
$(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})\cdot\boldsymbol{\mathit{E}}\cdot\boldsymbol{\mathit{F}}$
again from Remark 4.6, this gives

$(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})\cdot\boldsymbol{\mathit{E}}=96\mathsf{fib}.$

Therefore $\delta{\rm
fib}=(128/96)(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})\cdot\boldsymbol{\mathit{E}}=4(\boldsymbol{\mathit{E}}^{2}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}})/3$.
Our formulas become

$c_{2}(TY)=4(\boldsymbol{\mathit{E}}^{2}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}})/3-\boldsymbol{\mathit{E}}^{2}=(\boldsymbol{\mathit{E}}^{2}+4\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}})/3,$

and

${\rm
td}(TY)=1-\boldsymbol{\mathit{E}}/2+(\boldsymbol{\mathit{E}}^{2}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}})/9$

which completes the calculation. ∎

Next we compute the triple intersections of divisor classes on $Y$.

######  Proposition 4.8.

The triple intersections of divisor classes on the degree $1$ modular spectral
covering $Y$ are:

$\boldsymbol{\mathit{F}}^{3}=32,\;\;\;\;\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}}^{2}=64,\;\;\;\;\boldsymbol{\mathit{E}}^{2}\boldsymbol{\mathit{F}}=32,\;\;\;\;\boldsymbol{\mathit{E}}^{3}=-128.$

Recall that $H$ denotes the divisor class of the hyperplane section on $X$,
and $\boldsymbol{\mathit{F}}:=f^{*}(H)$. We have

$\displaystyle H^{1,1}(Y)\cap H^{2}(Y,{\mathbb{Q}})$
$\displaystyle=\langle\boldsymbol{\mathit{E}},\boldsymbol{\mathit{F}}\rangle,\text{
and }$ $\displaystyle H^{2,2}(Y)\cap H^{4}(Y,{\mathbb{Q}})$
$\displaystyle=\langle\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}},\boldsymbol{\mathit{F}}^{2}\rangle.$

The corresponding groups on $X$ are generated (over ${\mathbb{Q}}$) by $H$ and
$H^{2}$ respectively.

######  Lemma 4.9.

The map $f:Y\rightarrow X$ has degree $8$.

###### Proof.

Since over the very stable locus the map
$\pi_{*}(-):{\mathcal{P}}\dashrightarrow X$ is a proper morphism (see e.g.
[PP21, Zel20, PN20] it suffices to compute the number of preimages of a very
stable point in $X$ under $f:Y\to X$. Choose a general hence very stable point
$E\in X$. The fiber of the cotangent bundle $T^{\vee}_{E}X$ is the space
$H^{0}(\operatorname{End}_{0}(E)\otimes\omega_{C})$ of $\omega_{C}$-twisted
endomorphisms. The Hitchin base is the $3$-dimensional space
$H^{0}(C,\omega_{C}^{\otimes 2})\cong{\mathbb{A}}^{3}$ of quadratic
differentials on $C$. The map $T^{\vee}_{E}X\rightarrow{\mathbb{A}}^{3}$ is
the restriction of the Hitchin map and is thus given by three quadratic forms,
so the inverse image of a general point $\mathsf{b}\in{\mathbb{A}}^{3}$ is the
intersection of three quadrics: it has $8$ points. As
${\mathcal{P}}=\boldsymbol{h}^{-1}(\mathsf{b})$, and the exceptional divisor
$\boldsymbol{\mathit{E}}$ maps to the wobbly locus, we have that the
intersection of $Y\cap T^{\vee}_{E}X$ is equal to ${\mathcal{P}}\cap
T^{\vee}_{E}X$ and so is this set of $8$ points. Thus for a very stable $E$
and a generic ${\widetilde{C}}$ the inverse image $f^{-1}(E)\subset Y$
consists of $8$ points. ∎

######  Remark 4.10.

Note that the proof of Lemma 4.9 repeats verbatim to show that in the degree
$0$ case the map $f_{0}:Y_{0}\to X_{0}$ has degree $8$ as well.

We have $H^{3}=4$ since $X\subset{\mathbb{P}}^{5}$ is a degree $4$ subvariety.
Hence,

$\boldsymbol{\mathit{F}}^{3}=32.$

We also checked that
${\mathcal{O}}_{Y}(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})|_{\boldsymbol{\mathit{E}}}$
is a sum of $96$ fibers of the map $\boldsymbol{\mathit{E}}\to{\widehat{C}}$.
In particular, the divisor $(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})$
restricted to $\boldsymbol{\mathit{E}}$, has trivial self-intersection. We get
the formula

$(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})\cdot\boldsymbol{\mathit{E}}\cdot(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})=0,$

or

$\boldsymbol{\mathit{E}}^{3}+2\boldsymbol{\mathit{E}}^{2}\boldsymbol{\mathit{F}}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}}^{2}=0.$

######  Lemma 4.11.

The wobbly locus is the image ${\rm Wob}=f(\boldsymbol{\mathit{E}})\subset X$.
The intersection of ${\rm Wob}$ with a line $\ell$ in $X$ (in other words a
subvariety that is a line in ${\mathbb{P}}^{5}$) has $8$ points. The map
$f_{|\boldsymbol{\mathit{E}}}:\boldsymbol{\mathit{E}}\rightarrow{\rm Wob}$ is
generically $2$ to $1$, and if $D$ denotes the normalization of ${\rm Wob}$
(see Lemma 4.3 and Proposition 4.5 above) this map factors through a $2:1$ map
$\boldsymbol{\mathit{E}}\rightarrow D$.

###### Proof.

Recall from Lemma 4.3 that $D$ can be described as the space of downward or
outgoing directions to the curve ${\overline{C}}\subset\mathsf{Higgs}_{1}$. As
we saw in subsections 3.6 and 3.7 the incoming flow gives a $2:1$ map
${\widehat{C}}\rightarrow{\overline{C}}$ and a normal direction to a general
point of ${\widehat{C}}$ maps to a downward direction at the image point in
${\overline{C}}$. This gives the map $\boldsymbol{\mathit{E}}\rightarrow D$.
In the present case, downward flow lines are broken only once and the broken
flow line depends only on the first normal derivative to the point of
${\widehat{C}}$ in the Hitchin fiber, so the closure of the downward flow map
restricts on $\boldsymbol{\mathit{E}}$ to the composition
$\boldsymbol{\mathit{E}}\rightarrow D\rightarrow{\rm Wob}\subset X_{1}$. ∎

###### Proof of Proposition 4.8.

We saw above that $\boldsymbol{\mathit{F}}^{3}=32$.

A line $\ell$ lying on $X$ satisfies $\ell\cap H=1$, but also $(H^{2})\cap
H=4$; hence in the group $H^{2,2}(Y)\cap H^{4}(Y,{\mathbb{Q}})$ we have

$H^{2}=4\ell.$

This gives $H^{2}\cap{\rm Wob}=32$.

From the $2:1$ covering property we get

$\boldsymbol{\mathit{F}}^{2}\cap\boldsymbol{\mathit{E}}=H^{2}\cap 2{\rm Wob}$

so

$\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}}^{2}=64.$

Let us now calculate $\boldsymbol{\mathit{E}}^{3}$. It is the self-
intersection of the divisor class
$c_{1}(\mathcal{O}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}}))$ on the
surface $\boldsymbol{\mathit{E}}$. By the usual blowup picture the line bundle
$\mathcal{O}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}})$ is the
universal subbundle for the projectivization
$\mathbb{P}(N)=\boldsymbol{\mathit{E}}$ so the universal quotient bundle is

$UQ=b_{\boldsymbol{\mathit{E}}}^{*}(N)/{\mathcal{O}}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}}).$

In terms of divisor classes on $\boldsymbol{\mathit{E}}$ this gives

$c_{1}(UQ)=b_{\boldsymbol{\mathit{E}}}^{*}c_{1}(N)-c_{1}(\mathcal{O}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}}).$

The relative tangent bundle $T(\boldsymbol{\mathit{E}}/{\widehat{C}})$ of the
map $b_{\boldsymbol{\mathit{E}}}:\boldsymbol{\mathit{E}}\to{\widehat{C}}$ is
given by

$T(\boldsymbol{\mathit{E}}/{\widehat{C}})={\rm
Hom}({\mathcal{O}}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}}),UQ),$

which gives

$c_{1}(T(\boldsymbol{\mathit{E}}/{\widehat{C}}))=c_{1}(UQ)-c_{1}({\mathcal{O}}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}}))=b_{\boldsymbol{\mathit{E}}}^{*}c_{1}(N)-2c_{1}({\mathcal{O}}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}}))$

on the level of divisor classes.

Since $\boldsymbol{\mathit{E}}\cong{\widehat{C}}\times{\mathbb{P}}^{1}$, the
Neron-Severi group of $\boldsymbol{\mathit{E}}$ is freely generated by two
classes $\hat{\mathfrak{c}}$ and $\mathsf{fib}$, where $\hat{\mathfrak{c}}$ is
the class of ${\widehat{C}}\times\operatorname{pt}$ and as before
$\mathsf{fib}$ is the class of $\operatorname{pt}\times{\mathbb{P}}^{1}$. In
terms of these classes we have

$b_{\boldsymbol{\mathit{E}}}^{*}c_{1}(N)=128\mathsf{fib}\qquad\text{and}\qquad
c_{1}(T(\boldsymbol{\mathit{E}}/{\widehat{C}}))=2\hat{\mathfrak{c}}.$

Therefore we get

$c_{1}(\mathcal{O}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}}))=\frac{1}{2}\left(b_{\boldsymbol{\mathit{E}}}^{*}c_{1}(N)-c_{1}(T(\boldsymbol{\mathit{E}}/{\widehat{C}}))\right)=64\mathsf{fib}-\hat{\mathfrak{c}},$

and so taking into account that $\hat{\mathfrak{c}}^{2}=\mathsf{fib}^{2}=0$
and $\hat{\mathfrak{c}}\cdot\mathsf{fib}=1$ we get

$c_{1}(\mathcal{O}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}}))\cdot
c_{1}(\mathcal{O}_{\boldsymbol{\mathit{E}}}(\boldsymbol{\mathit{E}}))=(64\mathsf{fib}-\hat{\mathfrak{c}})^{2}=-128.$

That was the self-intersection on $\boldsymbol{\mathit{E}}$, so it gives on
$Y$, $\boldsymbol{\mathit{E}}^{3}=-128$.

Now from the formula
$\boldsymbol{\mathit{E}}^{3}+2\boldsymbol{\mathit{E}}^{2}\boldsymbol{\mathit{F}}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}}^{2}=0$
we get

$(-128)+2\boldsymbol{\mathit{E}}^{2}\boldsymbol{\mathit{F}}+(64)=0,\mbox{ so
}\boldsymbol{\mathit{E}}^{2}\boldsymbol{\mathit{F}}=(128-64)/2=32.$

This completes our list of intersection numbers on $Y$. ∎

###  4.3 Main construction

Let ${\mathcal{L}}_{0}$ be the pullback to $Y$ of a degree zero line bundle on
${\mathcal{P}}$, and set

${\mathcal{L}}_{a,b}:={\mathcal{L}}_{0}(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})\
\text{and}\ {\mathcal{V}}_{a,b}:=f_{*}({\mathcal{L}}_{a,b}).$

Following the notation we adopted in the definition of crude parabolic
structures in section 3.10 we will also set

${\mathcal{L}}^{\prime}_{a,b}:={\mathcal{L}}_{0}(a\boldsymbol{\mathit{F}}+b\boldsymbol{\mathit{E}})\
\subset\ {\mathcal{L}}_{a,b}\ \text{and}\
{\mathcal{V}}^{\prime}_{a,b}=f_{*}({\mathcal{L}}^{\prime}_{a,b})\ \subset\
{\mathcal{V}}_{a,b}.$

The notations are chosen so that the spectral line bundle defined in section
3.10 is precisely
${\mathcal{L}}^{\prime}={\mathcal{L}}_{0}(a\boldsymbol{\mathit{F}}+b\boldsymbol{\mathit{E}})$.
Notice that ${\mathcal{V}}^{\prime}_{a,b}={\mathcal{V}}_{a,b-1}$.

The next step is to look at the calculations of the Chern characters of the
direct image ${\mathcal{V}}_{a,b}=f_{*}({\mathcal{L}}_{a,b})$. The Chern
character of ${\mathcal{L}}_{a,b}$ is

${\rm
ch}({\mathcal{L}}_{a,b})=1+(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})+(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})^{2}/2.$

Putting the Todd class formula of Proposition 4.7 into the Grothendieck-
Riemann-Roch formula gives

$\displaystyle{\rm ch}({\mathcal{V}}_{a,b})$ $\displaystyle={\rm
ch}(f_{*}({\mathcal{L}}_{a,b}))$
$\displaystyle=f_{*}(1+(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})$
$\displaystyle\hskip
36.135pt+(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})^{2}/2)(1-\boldsymbol{\mathit{F}}+5\boldsymbol{\mathit{F}}^{2}/12)(1-\boldsymbol{\mathit{E}}/2+(\boldsymbol{\mathit{E}}^{2}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}})/9)$
$\displaystyle=f_{*}\left[1+((a-1)\boldsymbol{\mathit{F}}+(b+1/2)\boldsymbol{\mathit{E}})+(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})^{2}/2\right.$
$\displaystyle\hskip
36.135pt\left.+5\boldsymbol{\mathit{F}}^{2}/12+(\boldsymbol{\mathit{E}}^{2}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}})/9+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}}/2-(\boldsymbol{\mathit{F}}+\boldsymbol{\mathit{E}}/2)(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})\right].$

The rational cohomology of $X_{1}$ is generated by $H$ in degree $2$ and
$H^{2}$ in degree $4$, so if $a_{1}\in H^{2}(Y)$ and $a_{2}\in H^{4}(Y)$ are
classes on $Y$ we can write

$f_{*}(1+a_{1}+a_{2})=8+b_{1}H+b_{2}H^{2},$

for some rational numbers $b_{1},b_{2}\in\mathbb{Q}$. Then, using $H^{3}=4$ we
get $4b_{1}=H^{2}\cdot f_{*}(a_{1})=\boldsymbol{\mathit{F}}^{2}a_{1}$ and
$4b_{2}=H\cdot f_{*}(a_{2})=\boldsymbol{\mathit{F}}a_{2}$.

We get, using the calculations of Proposition 4.8:

$H^{2}\cdot{\rm
ch}_{1}({\mathcal{V}}_{a,b})=\boldsymbol{\mathit{F}}^{2}((a-1)\boldsymbol{\mathit{F}}+(b+1/2)\boldsymbol{\mathit{E}})=32(a-1)+64(b+1/2)=32a+64b$

so

${\rm ch}_{1}({\mathcal{V}}_{a,b})=(8a+16b)H.$

And

$\displaystyle H\cdot{\rm ch}_{2}({\mathcal{V}}_{a,b})$
$\displaystyle=\boldsymbol{\mathit{F}}\left[(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})^{2}/2\right.$
$\displaystyle\ \hskip
36.135pt\left.+5\boldsymbol{\mathit{F}}^{2}/12+(\boldsymbol{\mathit{E}}^{2}+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}})/9+\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}}/2-(\boldsymbol{\mathit{F}}+\boldsymbol{\mathit{E}}/2)(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})\right]$
$\displaystyle=(a^{2}/2+5/12-a)\boldsymbol{\mathit{F}}^{3}+(a(b+1)+1/9+1/2-(b+1+a/2))\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}}^{2}$
$\displaystyle\ \hskip
36.135pt+((b+1)^{2}/2+1/9-(b+1)/2)\boldsymbol{\mathit{E}}^{2}\boldsymbol{\mathit{F}}$
$\displaystyle=16a^{2}-32a+32a+64ab+16b^{2}+-64b+16b+40/3+64/9-32+32/9$
$\displaystyle=16a^{2}+16b^{2}+64ab-48b-8.$

We get

${\rm ch}_{2}({\mathcal{V}}_{a,b})=4a^{2}+16ab+4b^{2}-12b-2.$

Together these prove the following:

######  Proposition 4.12.

For ${\mathcal{L}}_{a,b}$ in the numerical class of
${\mathcal{O}}_{Y}(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})$ the
direct image ${\mathcal{V}}_{a,b}=\pi_{*}({\mathcal{L}}_{a,b})$ has Chern
character (truncated to codimension $2$ as usual)

${\rm ch}({\mathcal{V}}_{a,b})=8+(8a+16b)H+(4a^{2}+16ab+4b^{2}-12b-2)H^{2}.$

Later we will consider a parabolic modification of ${\mathcal{L}}_{a,b}$ over
the divisor $\boldsymbol{\mathit{E}}$, giving a new smooth parabolic structure
defined locally upstairs over the modular spectral cover or somewhat
equivalently over the projectivization of the bundle.

###  4.4 The cusp locus of the wobbly divisor

In this subsection we prove the part of Proposition 4.5 about the cuspidal
locus. Specifically we compute the locus $D_{\rm cusp}\subset D$ on which the
map $\boldsymbol{\mathsf{f}}:D\to X$ is not immersive. Recall that this map is
induced from the map $f:Y\to X$. More precisely, the restriction
$f_{|\boldsymbol{\mathit{E}}}:\boldsymbol{\mathit{E}}\to{\rm Wob}\subset X$
factors through the double cover
$\boldsymbol{\mathit{E}}={\widehat{C}}\times{\mathbb{P}}^{1}\to{\overline{C}}\times{\mathbb{P}}^{1}=D$
and $\boldsymbol{\mathsf{f}}:D\to X$ is just the induced map.

Let $(L,\eta)\in D={\overline{C}}\times\mathbb{P}^{1}$ be a point, and let
$E=\boldsymbol{\mathsf{f}}(L,\eta)\in X$ be the corresponding rank two bundle.
Our goal is to understand when the differential

$d\boldsymbol{\mathsf{f}}_{(L,\eta)}:T_{D,(L,\eta)}\to T_{X,E}$

has a non-trivial kernel.

Recall that $L\in{\overline{C}}\subset\operatorname{Jac}^{0}(C)$ means that
$L^{\otimes 2}(\boldsymbol{\mathsf{p}})$ is effective, i.e. that $L^{\otimes
2}\cong\mathcal{O}_{C}(t-\boldsymbol{\mathsf{p}})$ for some point $t\in C$.
Thus $L^{\otimes
2}(-\boldsymbol{\mathsf{p}})\cong\mathcal{O}_{C}(t-2\boldsymbol{\mathsf{p}})=\mathcal{O}_{C}(-t^{\prime})$,
where $t^{\prime}\in C$ is the point corresponding to $t$ under the
hyperelliptic involution.

Hence we have

$\operatorname{Ext}^{1}(L^{\vee}(\boldsymbol{\mathsf{p}}),L)=H^{1}(C,L^{\otimes
2}(-\boldsymbol{\mathsf{p}}))=H^{1}(C,\mathcal{O}_{C}(-t^{\prime})).$

From the long exact sequence in cohomology associated with the short exact
sequence
$0\to\mathcal{O}_{C}(-t^{\prime})\to\mathcal{O}_{C}\to\mathcal{O}_{t^{\prime}}\to
0$ it follows that the natural map

$\textstyle{\iota_{t^{\prime}}:}$$\textstyle{H^{1}(C,\mathcal{O}_{C}(-t^{\prime}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{H^{1}(C,\mathcal{O}_{C})}$

is an isomorphism, and hence we can view $\eta$ as an element in
$\iota_{t^{\prime}}^{-1}(\eta)\in\text{Ext}^{1}(L^{\vee}(\boldsymbol{\mathsf{p}}),L)$.
The rank two vector bundle $E=f(L,\eta)$ is the extension

$0\to L\to E\to L^{\vee}(\boldsymbol{\mathsf{p}})\to 0$

given by the extension class $\iota_{t^{\prime}}^{-1}(\eta)$.

We have $T_{D,(L,\eta)}=T_{{\overline{C}},L}\oplus T_{\mathbb{P}^{1},\eta}$
and so to understand the map $df_{(L,\eta)}$ it is enough to understand its
restrictions to the two coordinate lines $T_{{\overline{C}},L}\oplus\\{0\\}$
and $\\{0\\}\oplus T_{\mathbb{P}^{1},\eta}$ in $T_{D,(L,\eta)}$. These
restrictions which will abbreviate as

$\textstyle{d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{{\overline{C}},L}}:}$$\textstyle{T_{{\overline{C}},L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{X,E}}$$\textstyle{d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{\mathbb{P}^{1},\eta}}:}$$\textstyle{T_{\mathbb{P}^{1},\eta}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{X,E}}$

have a natural modular interpretation.

####  4.4.1 Interpretation of
$d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{\mathbb{P}^{1},\eta}}$

Fix $L\in{\overline{C}}$ and hence the points $t,t^{\prime}\in C$. The
extensions of $L^{\vee}(\boldsymbol{\mathsf{p}})$ by $L$ corresponding to a
varying point $\eta\in\mathbb{P}(H^{1}(C,\mathcal{O}_{C}))$ fit in a natural
universal family parametrized by $\mathbb{P}^{1}$. Indeed, consider the
surface $C\times\mathbb{P}^{1}$. By Künneth we have

$\displaystyle\text{Ext}^{1}_{C\times\mathbb{P}^{1}}(p_{C}^{*}L^{\vee}(\boldsymbol{\mathsf{p}}),p_{C}^{*}L\otimes
p_{\mathbb{P}^{1}}^{*}\mathcal{O}(1))$
$\displaystyle=H^{1}(C\times\mathbb{P}^{1},p_{C}^{*}\mathcal{O}(-t^{\prime})\otimes
p_{\mathbb{P}^{1}}^{*}\mathcal{O}(1))$
$\displaystyle=H^{1}(C,\mathcal{O}(-t^{\prime}))\otimes
H^{0}(\mathbb{P}^{1},\mathcal{O}(1))$ $\displaystyle\cong
H^{1}(C,\mathcal{O})\otimes H^{1}(C,\mathcal{O})^{\vee}.$

Here in the last step we used $\iota_{t^{\prime}}$ to identify
$H^{1}(C,\mathcal{O}(-t^{\prime}))$ with $H^{1}(C,\mathcal{O})$ and we used
$\mathbb{P}^{1}=\mathbb{P}(H^{1}(C,\mathcal{O}))$ to identify
$H^{0}(\mathbb{P}^{1},\mathcal{O}(1))$ with $H^{1}(C,\mathcal{O})^{\vee}$.

The extension

$0\to p_{C}^{*}L\otimes p_{\mathbb{P}^{1}}^{*}\mathcal{O}(1)\to\mathscr{E}\to
p_{C}^{*}L^{\vee}(\boldsymbol{\mathsf{p}})\to 0$

corresponding to the identity element in $H^{1}(C,\mathcal{O})\otimes
H^{1}(C,\mathcal{O})^{\vee}$ is a rank two bundle $\mathscr{E}$ on
$C\times\mathbb{P}^{1}$ whose restrictiion to $C\times\\{\xi\\}$ is the
extension

$0\to L\to\mathscr{E}_{\xi}\to L^{\vee}(\boldsymbol{\mathsf{p}})\to 0$

given by the class $\iota_{t^{\prime}}^{-1}(\xi)$. In particular we have
$\mathscr{E}_{\xi}=E$.

Therefore
$\boldsymbol{\mathsf{f}}_{|\\{L\\}\times\mathbb{P}^{1}}:\mathbb{P}^{1}\to X$
can be viewed as the classifying map for the bundle $\mathscr{E}\to
C\times\mathbb{P}^{1}$. Let $\text{End}_{0}(E)$ denote the bundle of traceless
endomorphisms of $E$. By deformation theory we have
$T_{X,E}=H^{1}(C,\text{End}_{0}(E))$ and we can identify the restricted map

$d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{\mathbb{P}^{1},\eta}}:T_{\mathbb{P}^{1},\eta}\to
H^{1}(C,\text{End}_{0}(E))$

with the Kodaira-Spencer class of the family $\mathscr{E}\to
C\times\mathbb{P}^{1}$ at $\eta$. But tensoring a family with a fixed line
bundle does not change the Kodaira-Spencer class and so equivalently we can
compute the Kodaira-Spencer class for the family
$\mathscr{F}=\mathscr{E}\otimes p_{C}^{*}L(-\boldsymbol{\mathsf{p}})\to
C\times\mathbb{P}^{1}$. By definition $\mathscr{F}$ is the extension

$0\to\mathcal{O}(-t^{\prime})\boxtimes\mathcal{O}(1)\to\mathscr{F}\to\mathcal{O}_{C\times\mathbb{P}^{1}}\to
0$ (19)

corresponding to the element

$\iota_{t^{\prime}}^{-1}\in H^{1}(C,\mathcal{O}(-t^{\prime}))\otimes
H^{1}(C,\mathcal{O}))^{\vee}=\text{Ext}^{1}_{C\times\mathbb{P}^{1}}(\mathcal{O},\mathcal{O}(-t^{\prime})\boxtimes\mathcal{O}(1)),$

and so is the universal family of extensions

$0\to\mathcal{O}_{C}(-t^{\prime})\to\mathscr{F}_{\xi}\to\mathcal{O}_{C}\to 0,$

corresponding to points $\xi\in\mathbb{P}^{1}$. To simplify notation we will
denote $\mathscr{F}_{\eta}$ by $F$.

Now from the exact sequence (19) it follows that
$\text{End}_{0}(\mathscr{F})=\text{End}_{0}(\mathscr{E})$ maps naturally onto
$\text{Hom}(\mathcal{O}(-t^{\prime})\boxtimes\mathcal{O}(1),\mathscr{F})$ and
so we have an exact sequence

$0\to\mathcal{O}(-t^{\prime})\boxtimes\mathcal{O}(1)\to\text{End}_{0}(\mathscr{F})\to\mathscr{F}\otimes(\mathcal{O}(t^{\prime})\boxtimes\mathcal{O}(-1))\to
0$

which for each $\xi\in\mathbb{P}^{1}$ specializes to a short exact sequence

$0\to\mathcal{O}_{C}(-t^{\prime})\to\text{End}_{0}(\mathscr{E}_{\xi})\to\mathscr{F}_{\xi}(t^{\prime})\to
0.$ (20)

Since the family $\mathscr{E}$ encodes the variation of the bundle
$\mathscr{E}_{\xi}$ as the extension class $\xi\in H^{1}(C,\mathcal{O}))$
varies, the Kodaira-Spencer class
$\mathsf{ks}^{\mathscr{E}}_{\xi}:T_{\mathbb{P}^{1},\xi}\to
H^{1}(C,\text{End}_{0}(\mathscr{E}_{\xi}))$ is captured by the maps between
cohomology of the terms in (20). Indeed, taking into account that
$H^{0}(C,\mathcal{O}(-t^{\prime}))=0$ and
$H^{0}(C,\text{End}_{0}(\mathscr{E}_{\xi}))=0$, the long exact sequence of
cohomology associated with (20) reads

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{H^{0}(C,\mathscr{F}(t^{\prime}))}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{H^{1}(C,\mathcal{O}(-t^{\prime}))}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{H^{1}(C,\text{End}_{0}(\mathscr{E}_{\xi}))}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{H^{1}(C,\mathscr{F}_{\xi}(t^{\prime}))}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$
(21)

Also, in the long exact sequence associated with

$0\to\mathcal{O}_{C}\to\mathscr{F}_{\xi}(t^{\prime})\to\mathcal{O}_{C}(t^{\prime})\to
0$

the first edge homomorphism $H^{0}(C,\mathcal{O}(t^{\prime}))\to
H^{1}(C,\mathcal{O})$ is given simply by cup product with
$\iota_{t^{\prime}}^{-1}(\xi)\in H^{1}(C,\mathcal{O}(-t^{\prime}))$. Since
$\xi$ is not zero and $H^{0}(C,\mathcal{O}(t^{\prime}))\cong\mathbb{C}$ this
implies that this edge homomorphism is injective and so the natural map
$H^{0}(C,\mathcal{O})\to H^{0}(C,\mathscr{F}_{\xi}(t^{\prime}))$ is an
isomorphism. If we identify $H^{0}(C,\mathscr{F}_{\xi}(t^{\prime}))$ with
$H^{0}(C,\mathcal{O})$ via this isomorphism, then the first edge homomorphism
in (21) becomes the cup product

$\iota_{q^{\prime}}^{-1}(\xi)\cup\ :H^{0}(C,\mathcal{O})\to
H^{1}(C,\mathcal{O}(-t^{\prime})).$

Using $\iota_{t^{\prime}}$ to identify $H^{1}(C,\mathcal{O}(-t^{\prime}))$
with $H^{1}(C,\mathcal{O})$ we then see that the long exact sequence (21)
reduces to a short exact sequence

$0\to
H^{1}(C,\mathcal{O})/\mathbb{C}\cdot\xi\to{H^{1}(C,\text{End}_{0}(\mathscr{E}_{\xi}))}\to{H^{1}(C,\mathscr{F}_{\xi}(t^{\prime}))}\to
0.$

But $H^{1}(C,\mathcal{O})/\mathbb{C}\cdot\xi=T_{\mathbb{P}^{1},\xi}$ and the
first map in this sequence by construction sends an infinitesimal deformation
of the extension $\xi$ to an infinitesimal deformation of $\mathscr{E}_{\xi}$.
Therefore this map is the Kodaira-Spencer map and we can rewrite the previous
sequence as

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{\mathbb{P}^{1},\xi}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathsf{ks}^{\mathscr{E}}_{\xi}}$$\textstyle{{H^{1}(C,\text{End}_{0}(\mathscr{E}_{\xi}))}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{H^{1}(C,\mathscr{F}_{\xi}(t^{\prime}))}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

or in the special case $\xi=\eta$ as

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T_{\mathbb{P}^{1},\eta}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{\mathbb{P}^{1},\eta}}}$$\textstyle{{H^{1}(C,\text{End}_{0}(E))}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{H^{1}(C,F(t^{\prime}))}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

This gives the desired modular interpretation of
$d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{\mathbb{P}^{1},\eta}}$, and confirms
that $\boldsymbol{\mathsf{f}}$ restricted to any ruling
$\\{L\\}\times\mathbb{P}^{1}$ is an immersion.

####  4.4.2 Interpretation of
$d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{{\overline{C}},L}}$

Fix $\eta\in\mathbb{P}^{1}=\mathbb{P}(H^{1}(C,\mathcal{O}))$. The extensions
of $L^{\vee}(\boldsymbol{\mathsf{p}})$ by $L$ corresponding to the fixed
extension class $\eta$ and a variable $L\in{\overline{C}}$ fit in a natural
family parametrized by ${\overline{C}}$. Indeed, consider the surface $C\times
C$ with its diagonal divisor $\Delta\subset C\times C$. Pushing down the short
exact sequence

$0\to\mathcal{O}(-\Delta)\to\mathcal{O}\to\mathcal{O}_{\Delta}\to 0$

via the projection $p_{2}:C\times C\to C$ onto the second factor gives a long
exact sequence of direct images:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{p_{2*}\mathcal{O}(-\Delta)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{p_{2*}\mathcal{O}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{p_{2*}\mathcal{O}_{\Delta}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{R^{1}p_{2*}\mathcal{O}(-\Delta)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{R^{1}p_{2*}\mathcal{O}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{R^{1}p_{2*}\mathcal{O}_{\Delta}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

We have $p_{2*}\mathcal{O}(-\Delta)=0$ since $\mathcal{O}(-\Delta)$ has
negative degree along the fibers of $p_{2}$. Since $p_{2}:\Delta\to C$ is an
isomorphism we also have $p_{2*}\mathcal{O}_{\Delta}=\mathcal{O}_{C}$ and
$R^{1}p_{2*}\mathcal{O}_{\Delta}=0$. Finally by Künneth we have
$p_{2*}\mathcal{O}=\mathcal{O}_{C}$ and
$R^{1}p_{2*}\mathcal{O}=H^{1}(C,\mathcal{O})\otimes\mathcal{O}_{C}$.
Substituting these in the long exact sequence gives

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathcal{O}_{C}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathcal{O}_{C}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{R^{1}p_{2*}\mathcal{O}(-\Delta)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{H^{1}(C,\mathcal{O})\otimes\mathcal{O}_{C}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

Therefore we have

$p_{2*}\mathcal{O}(-\Delta)=0,\qquad\qquad
R^{1}p_{2*}\mathcal{O}(-\Delta)=H^{1}(C,\mathcal{O})\otimes\mathcal{O}_{C}$

With this in mind we compute

$\displaystyle\text{Ext}^{1}_{C\times C}(\mathcal{O},\mathcal{O}(-\Delta))$
$\displaystyle=H^{1}(C\times C,\mathcal{O}(-\Delta))$
$\displaystyle=H^{0}(C,R^{1}p_{1*}\mathcal{O}(-\Delta))$
$\displaystyle=H^{1}(C,\mathcal{O})\otimes H^{0}(C,\mathcal{O})$

where at the second step we used the Leray spectral sequence for $p_{2}$.

Therefore the class $\eta\otimes 1\in H^{1}(C,\mathcal{O})\otimes
H^{0}(C,\mathcal{O})$ corresponds to a rank two bundle $\mathbb{F}$ on
$C\times C$ which is an extension

$0\to\mathcal{O}_{C\times C}(-\Delta)\to\mathbb{F}\to\mathcal{O}_{C\times
C}\to 0.$

By construction, for every point $x\in C$, the restriction
$\mathbb{F}_{x}=\mathbb{F}_{|C\times\\{x\\}}$ is the extension

$0\to\mathcal{O}_{C}(-x)\to\mathbb{F}_{x}\to\mathcal{O}_{C}\to 0$

given by the class $\iota_{x}^{-1}(\eta)\in H^{1}(C,\mathcal{O}(-x))$. In
particular we have $\mathbb{F}_{t^{\prime}}=F$.

Now consider the degree $16$ map $C\times{\overline{C}}\to C\times C$ which is
the identity on the first factor and on the second factor is the composition
of the natural map ${\overline{C}}\to C$ with the hyperelliptic
involution555The canonical map from ${\overline{C}}$ to $C$ assigns to each
$L$ the unique point $t\in C$ such that $L^{\otimes 2}=\mathcal{O}_{C}(t)$.
The map we consider here sends $L$ not to $t$ but to its image under the
hyperelliptic involution $t^{\prime}$. Let $\mathscr{F}\to
C\times{\overline{C}}$ be the pull back of $\mathbb{F}$ by this map, and let
$\mathscr{L}\to C\times{\overline{C}}$ be the Poincaré line bundle. Then the
bundle
$\mathscr{E}=\mathscr{F}\otimes\mathscr{L}^{\vee}(\\{\boldsymbol{\mathsf{p}}\\}\times{\overline{C}})$
is an extension

$0\to\mathscr{L}\to\mathscr{E}\to\mathscr{L}^{\vee}(\\{\boldsymbol{\mathsf{p}}\\}\times{\overline{C}})\to
0.$

For each $M\in{\overline{C}}$ restricts to the bundle
$\mathscr{E}_{M}:=\mathscr{E}_{|C\times\\{M\\}}$ which is the extension

$0\to M\to\mathscr{E}_{M}\to M^{\vee}(\boldsymbol{\mathsf{p}})\to 0$

corresponding to the class $\eta$. In particular $\mathscr{E}_{L}=E$ and the
map
$\boldsymbol{\mathsf{f}}_{|{\overline{C}}\times\\{\eta\\}}:{\overline{C}}\to
X$ is the classifying map for the family $\mathscr{E}\to
C\times{\overline{C}}$.

This identifies the restricted differential
$d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{{\overline{C}},L}}$ with the Kodaira-
Spencer class

$\mathsf{ks}^{\mathscr{E}}_{L}:T_{{\overline{C}},L}\to
H^{1}(C,\text{End}_{0}(E))$

of the family $\mathscr{E}$ computed $L\in{\overline{C}}$.

The family $\mathscr{E}$ differs from the family of bundles $\mathscr{F}$ by a
tensoring with a family of line bundles. So, even though $\mathscr{F}$ is a
family of bundles with varying determinant, the traceless parts of the
Kodaira-Spencer classes for $\mathscr{E}$ and $\mathscr{F}$ are equal. Given a
family of rank two bundles $\mathcal{F}$ on $C$ let us write
$\kappa^{\mathcal{F}}=\mathsf{ks}^{\mathcal{F}}-(\text{tr}(\mathsf{ks}^{\mathcal{F}})/2)\cdot\text{id}$
for the traceless part of the Kodaira-Spencer class of $\mathcal{F}$. With
this notation we then have

$d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{{\overline{C}},L}}=\mathsf{ks}^{\mathscr{E}}_{L}=\kappa^{\mathscr{E}}_{L}=\kappa^{\mathscr{F}}_{L}.$

As a last simplification, note that the map ${\overline{C}}\to C$, $L\mapsto
t^{\prime}$ is étale and so its differential is an isomorphism
$T_{{\overline{C}},L}\stackrel{{\scriptstyle\cong}}{{\to}}T_{C,t^{\prime}}$ of
tangent spaces. Taking into account that the family $\mathscr{F}$ is the
pullback of the family $\mathbb{F}$ we conclude that modulo the isomorphism
$T_{{\overline{C}},L}\stackrel{{\scriptstyle\cong}}{{\to}}T_{C,t^{\prime}}$
the traceless Kodaira-Spencer class $\kappa^{\mathscr{F}}_{L}$ equals the
traceless Kodaira-Spencer class $\kappa^{\mathbb{F}}_{t^{\prime}}$.

####  4.4.3 The kernel of the differential

After these preliminaries we are now ready to understand where the map
$\boldsymbol{\mathsf{f}}:D\to X$ is not immersive.

######  Proposition 4.13.

Let $(L,\eta)\in D={\overline{C}}\times\mathbb{P}^{1}$ be a point, and let
$E=\boldsymbol{\mathsf{f}}(L,\eta)\in X$ be the corresponding rank two bundle.
Then the differential

$d\boldsymbol{\mathsf{f}}_{(L,\eta)}:T_{D,(L,\eta)}\to T_{X,E}$

has a non-trivial kernel if and only if $L\in{\overline{C}}$ maps to
$\eta\in\mathbb{P}^{1}=\mathbb{P}(H^{1}(C,\mathcal{O}_{C}))$ under the natural
degree $32$ map ${\overline{C}}\to C\to\mathbb{P}^{1}$.

Proof. Since $d\boldsymbol{\mathsf{f}}_{(L,\eta)}$ is determined by its
restricions to $T_{{\overline{C}},L}$ and $T_{\mathbb{P}^{1},\eta}$ and since
we saw that $d\boldsymbol{\mathsf{f}}_{(L,\eta)|T_{\mathbb{P}^{1},\eta}}$ is
always injective, it suffices to characterize all points $(L,\eta)\in D$ for
which the image $d\boldsymbol{\mathsf{f}}_{(L,\eta)}(T_{{\overline{C}},L})$ is
contained in the line
$d\boldsymbol{\mathsf{f}}_{(L,\eta)}(T_{\mathbb{P}^{1},\eta})\subset
T_{X,E}=H^{1}(C,\text{End}_{0}(E))=H^{1}(C,\text{End}_{0}(F))$.

In view of the modular interpretations of the restricted differentials in
sections 4.4.1 and 4.4.2 this question is equivalent to the problem of
characterizing all points $(t^{\prime},\eta)\in C\times\mathbb{P}^{1}$ such
that

$\kappa^{\mathbb{F}}_{t^{\prime}}(T_{C,t^{\prime}})\subset\ker\left[H^{1}(C,\text{End}_{0}(F))\to
H^{1}(C,F(t^{\prime}))\right].$ (22)

Here as before $\mathbb{F}$ is the extension

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}_{C\times
C}(-\Delta)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\textstyle{\mathbb{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{b}$$\textstyle{\mathcal{O}_{C\times
C}\to 0,}$ (23)

corresponding to $\eta$ and
$F=\mathbb{F}_{t^{\prime}}:=\mathbb{F}_{|C\times\\{t^{\prime}\\}}$.

More invariantly the condition (22) can be rewritten as follows. The short
exact sequence (23) induces a short exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\text{Hom}(\mathcal{O},\mathcal{O}(-\Delta))\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a\circ-\circ
b}$$\textstyle{\text{End}_{0}(\mathbb{F})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{-\circ
a}$$\textstyle{\text{Hom}(\mathcal{O}(-\Delta),\mathbb{F})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{\mathcal{O}(-\Delta)}$$\textstyle{\mathbb{F}(\Delta)}$
(24)

Pushing this down the map $-\circ
a:\text{End}_{0}(\mathbb{F})\to\mathbb{F}(\Delta)$ via the second projection
$p_{2}:C\times C\to C$ gives us a map of coherent sheaves

$R^{1}p_{2*}(-\circ a):R^{1}p_{2*}\text{End}_{0}(\mathbb{F})\to
R^{1}p_{2*}\mathbb{F}(\Delta).$

We also have the universal traceless Kodaira-Spencer class of $\mathbb{F}$

$\kappa^{\mathbb{F}}:T_{C}\to R^{1}p_{2*}\text{End}_{0}(\mathbb{F}).$

Composing these two maps we get a map

$R^{1}p_{2*}(-\circ a)\circ\kappa^{\mathbb{F}}:T_{C}\to
R^{1}p_{2*}\mathbb{F}(\Delta),$

or equivalently a global section

$R^{1}p_{2*}(-\circ a)\circ\kappa^{\mathbb{F}}\in
H^{0}\left(C,\Omega^{1}_{C}\otimes R^{1}p_{2*}\mathbb{F}(\Delta)\right).$ (25)

Note that the bundle $\mathbb{F}$ and the global section (25) depend only on
the class $\eta\in\mathbb{P}(H^{1}(C,\mathcal{O}))$. The condition (22) is
simply the condition that the point $t^{\prime}$ belongs to the zero locus of
this section. Therefore we need to describe all pairs $(t^{\prime},\eta)\in
C\times\mathbb{P}^{1}$ for which the section (25) defined by $\eta$ vanishes
at the point $t^{\prime}$.

To understand these pairs observe that $\mathcal{O}(-\Delta)\subset\mathbb{F}$
can be viewed as the family of filtered bundles
$\mathcal{O}(-x)\subset\mathbb{F}_{x}$ parametrized by $x\in C$. In
particular, the traceless Kodaira-Spencer class $\kappa^{\mathbb{F}}$ will
preserve the filtration. In other words if ${}^{\rm
filt}\text{End}_{0}(\mathbb{F})$ denotes the bundle of traceless endomorphisms
of $\mathbb{F}$ preserving the subbundle $\mathcal{O}(-\Delta)$, then the
global traceless Kodaira-Spencer class $\kappa^{\mathbb{F}}$ is induced from
the traceless part of the Kodaira-Spencer class

${}^{\rm filt}\kappa^{\mathbb{F}}:T_{C}\to R^{1}p_{2*}{}^{\rm
filt}\text{End}_{0}(\mathbb{F})$

classifying filtered deformations. This means that $\kappa^{\mathbb{F}}$
factors as

$\textstyle{T_{C}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{}^{\rm
filt}\kappa^{\mathbb{F}}}$$\scriptstyle{\kappa^{\mathbb{F}}}$$\textstyle{R^{1}p_{2*}{}^{\rm
filt}\text{End}_{0}(\mathbb{F})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{1}p_{2*}\text{End}_{0}(\mathbb{F}).}$

By definition

${}^{\rm
filt}\text{End}_{0}(\mathbb{F})=\left\\{\varphi\in\text{End}(F)\,\left|\,b\circ\varphi\circ
a=0\ \text{and}\ \text{tr}(\varphi)=0\right.\right\\},$

and hence

${}^{\rm filt}\text{End}_{0}(\mathbb{F})=\ker\left[\lx@xy@svg{\hbox{\raise
0.0pt\hbox{\kern
18.51115pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 1.0pt\raise
0.0pt\hbox{$\textstyle{\text{End}_{0}(\mathbb{F})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
19.49236pt\raise 5.43056pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.43056pt\hbox{$\scriptstyle{b\circ-\circ a}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 42.51115pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
42.51115pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
1.0pt\raise
0.0pt\hbox{$\textstyle{\mathcal{O}(\Delta)}$}}}}}}}\ignorespaces}}}}\ignorespaces\right].$

This implies that $R^{1}p_{2*}(-\circ a)\circ\kappa^{\mathbb{F}}$ maps $T_{C}$
into the subsheaf

$\ker\left[R^{1}p_{2*}\mathbb{F}(\Delta)\to
R^{1}p_{2*}\mathcal{O}(\Delta)\right]\subset
R^{1}p_{2*}\text{End}_{0}(\mathbb{F}.$

But from the long exact sequence of $p_{2}$ direct images of the short exact
sequence

$0\to\mathcal{O}\to\mathbb{F}(\Delta)\to\mathcal{O}(\Delta)\to 0$

it follows immediately that

$\ker\left[\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
18.51115pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 1.0pt\raise
0.0pt\hbox{$\textstyle{\text{End}_{0}(\mathbb{F})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
19.49236pt\raise 5.43056pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.43056pt\hbox{$\scriptstyle{b\circ-\circ a}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 42.51115pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
42.51115pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
1.0pt\raise
0.0pt\hbox{$\textstyle{\mathcal{O}(\Delta)}$}}}}}}}\ignorespaces}}}}\ignorespaces\right]=(H^{1}(C,\mathcal{O})/\eta)\otimes\mathcal{O}_{C}.$

This imlplies that the section $R^{1}p_{2*}(-\circ a)\circ\kappa^{\mathbb{F}}$
in $\omega_{C}\otimes R^{1}p_{2*}\mathbb{F}(\Delta)$ is in fact the
tautological section in the natural subbundle

$\omega_{C}\otimes\left((H^{1}(C,\mathcal{O})/\eta)\otimes\mathcal{O}_{C})\right)=(H^{1}(C,\mathcal{O})/\eta)\otimes\omega_{C},$

i.e. is the element in

$(H^{1}(C,\mathcal{O})/\eta)\otimes
H^{0}(C,\omega_{C})=(H^{1}(C,\mathcal{O})/\eta)\otimes
H^{1}(C,\mathcal{O})^{\vee}$

corresponding to the graph of the quotient map $H^{1}(C,\mathcal{O})\to
H^{1}(C,\mathcal{O})/\eta$. This is precisely the section of $\omega_{C}$ that
vanishes at the preimage of $\eta$ under the hyperelliptic map
$\mathsf{h}_{C}:C\to\mathbb{P}^{1}$. Therefore the set of pairs
$(t^{\prime},\eta)$ we were seeking is just the graph of the hyperelliptic map
and so the set of points $(L,\eta)$ at which $\boldsymbol{\mathsf{f}}$ is not
immersive is the graph of the projection ${\overline{C}}\to
C\to\mathbb{P}^{1}$. This completes the proof of the proposition. $\Box$

###  4.5 The trigonal cover of a general line

Recall from [New68, Theorem 2], [NR69, Theorem 5] that the variety of lines on
$X$ is isomorphic to the Jacobian $\operatorname{Jac}^{0}(C)$ of $C$. Taking
up the theme used in the proof of Lemma 4.4 in more detail, consider a general
line $\ell\subset X$. It corresponds to a degree $0$ line bundle
$A\in\operatorname{Jac}^{0}(C)$, and in terms of vector bundles the line is
the set of isomorphism classes of nonzero extensions of the form

$0\rightarrow A\rightarrow V\rightarrow
A^{\vee}(\boldsymbol{\mathsf{p}})\rightarrow 0.$

A point corresponding to a given extension, will lie on a different line
$\ell^{\prime}$ corresponding to $A^{\prime}\in\operatorname{Jac}^{0}(C)$
distinct from $A$, if and only if there is a map $A^{\prime}\rightarrow
A^{\vee}(\boldsymbol{\mathsf{p}})$ such that the pullback extension splits.
Such a map corresponds to a point $t\in C$ with
$A^{\prime}=A^{\vee}(\boldsymbol{\mathsf{p}}-t)$.

The splitting of the extension may be analyzed as follows. We have

${\rm Ext}^{1}(A^{\vee}(\boldsymbol{\mathsf{p}}),A)\cong H^{1}(C,A^{\otimes
2}(-\boldsymbol{\mathsf{p}}))$

which is dual to $H^{0}(C,A^{\otimes-2}(3\boldsymbol{\mathsf{p}}))$. These
spaces have dimension $2$, so a nonzero element of $H^{1}(C,A^{\otimes
2}(-\boldsymbol{\mathsf{p}}))$ modulo scalars also corresponds to a nonzero
element of $H^{0}(C,A^{\otimes-2}(3\boldsymbol{\mathsf{p}}))$ modulo scalars,
the unique element that pairs with it to zero by the duality.

The extension splits when pulled back to $A^{\vee}(\boldsymbol{\mathsf{p}}-t)$
if and only if the class maps to zero under the map

$H^{1}(C,A^{\otimes 2}(-\boldsymbol{\mathsf{p}}))\rightarrow
H^{1}(C,A^{\otimes 2}(t-\boldsymbol{\mathsf{p}})),$

or equivalently if the dual element is in the image of the map

$H^{0}(C,A^{\otimes-2}(3\boldsymbol{\mathsf{p}}-t))\rightarrow
H^{0}(C,A^{\otimes-2}(3\boldsymbol{\mathsf{p}})).$

In turn, this is equivalent to saying that our dual element, viewed as a
section of the line bundle $A^{\otimes-2}(3\boldsymbol{\mathsf{p}})$, vanishes
at $t$.

In conclusion, each extension corresponds to a class (up to scalar multiples)
of nonzero sections of $A^{\otimes-2}(3\boldsymbol{\mathsf{p}})$, and the
other lines $\ell^{\prime}$ passing through the point determined by the
extension correspond to the points where this section vanishes.

We may view this as saying that the complete linear system associated to the
line bundle $A^{\otimes-2}(3\boldsymbol{\mathsf{p}})$ of degree $3$ determines
a trigonal map $C\rightarrow\ell\cong{\mathbb{P}}^{1}$, and the other three
lines through a point on $\ell$ correspond to the fibers.

######  Proposition 4.14.

The trigonal cover ramifies over $8$ points in $\ell$. These are the $8$
intersection points of $\ell$ with the wobbly locus. If $x$ is such an
intersection point, and if we write the divisor in $C$ as $2u+v$ then $u$
corresponds to the line through that point in the wobbly locus.

###### Proof.

If $A$ is general then the line bundle
$A^{\otimes-2}(3\boldsymbol{\mathsf{p}})$ is general and hence has no base
points. For such an $A$ the trigonal map will be a morphism. The Hurwitz
formula implies that there are $8$ ramification points. A general fiber of the
trigonal cover is a section of $A^{\otimes-2}(3\boldsymbol{\mathsf{p}})$ that
vanishes at three points $u,v,w\in C$. A simple ramification is a point $u$
such that there is a section of $A^{\otimes-2}(3\boldsymbol{\mathsf{p}})$ that
vanishes at $2u+v$. We get
$A^{\otimes-2}(3\boldsymbol{\mathsf{p}})\cong{\mathcal{O}}_{C}(2u+v)$. As
discussed above, the extension corresponding to this point of $\ell$ splits on

$A^{-1}(\boldsymbol{\mathsf{p}}-u)\cong A(u+v-2\boldsymbol{\mathsf{p}})\cong
A(u-v^{\prime})$

where $v^{\prime}$ is the hyperelliptic conjugate of $v$, and we get a diagram

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A^{-1}(\boldsymbol{\mathsf{p}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{A(u-v^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

and the bundle fits in an extension

$0\rightarrow A(u-v^{\prime})\rightarrow E\rightarrow
A^{-1}(3\boldsymbol{\mathsf{p}}-u-v)\cong A(u)\rightarrow 0.$

Now
$A(u-v^{\prime})\otimes\omega_{C}=A(2\boldsymbol{\mathsf{p}}+u-v^{\prime})=A(u+v)$
so we get a nonzero map

$A(u)\rightarrow A(u+v)\cong A(u-v^{\prime})\otimes\omega_{C}$

yielding a nonzero nilpotent Higgs field. Therefore, the bundle $E$ associated
to the given extension is in the wobbly locus. ∎

_Remark:_ We already argued synthetically in section 2.5 that the line $\ell$
goes through the cuspidal locus of the wobbly locus when the divisor in $C$
has the form $3u$.

Choose a general line $\ell\subset X=X_{1}$. Over this line we have the
trigonal cover $C\rightarrow\ell$ associating to each point $x\in\ell$ the set
of three lines passing through $x$ that are distinct from $\ell$. Let $u,v,w$
be the points of $C$ over $x\in\ell$.

Let $Y_{\ell}:=Y\times_{X}\ell$. It is a degree $8$ ramified cover of $\ell$.

######  Lemma 4.15.

Assume $x$ general in $\ell$. The fiber of $Y_{\ell}\rightarrow\ell$ over the
point $x$ is naturally isomorphic to the set of $8$ points obtained by
choosing $\tilde{u},\tilde{v},\tilde{w}\in{\widetilde{C}}$ over $u,v,w\in C$
respectively.

###### Proof.

Let $L$ be the line bundle of degree $0$ on $C$ corresponding to the line
$\ell$, and let $E$ be the rank $2$ vector bundle corresponding to $x$. It
fits into an exact sequence

$0\rightarrow L\rightarrow E\rightarrow
L^{-1}(\boldsymbol{\mathsf{p}})\rightarrow 0$

whose extension class corresponds to the point in $\ell$. The fiber of $Y$
over $x$ is the set of line bundles $U$ on ${\widetilde{C}}$, of degree $3$,
such that $\pi_{*}(U)\cong E$. The map $L\rightarrow\pi_{*}(U)$ corresponds by
adjunction to a map $\pi^{*}(L)\rightarrow U$. We get

$U\cong\pi^{*}(L)\otimes{\mathcal{O}}_{{\widetilde{C}}}(\tilde{u}+\tilde{v}+\tilde{w}).$

Let $u,v,w$ be the images of $\tilde{u}$, $\tilde{v}$, $\tilde{w}$ in $C$. The
determinant of $\pi_{*}(U)$ is $L^{\otimes
2}\otimes{\mathcal{O}}_{C}(u+v+w)\otimes\omega_{C}^{-1}$ so we get

$L^{\otimes 2}(u+v+w)\cong{\mathcal{O}}_{C}(3\boldsymbol{\mathsf{p}}).$

It means that $u+v+w$ is in the linear system
$|L^{\otimes-2}(3\boldsymbol{\mathsf{p}})|$, in other words it is one of the
fibers of the trigonal cover of $\ell$. To show that it is the fiber over the
point corresponding to the given extension defining $E$, we need to see that
the extension splits when restricted to $L^{-1}(\boldsymbol{\mathsf{p}}-u)$
(and the same for $v,w$). For this, in turn, it suffices to see that there is
a nonzero map

$\pi^{*}(L^{-1}(\boldsymbol{\mathsf{p}}-u))\rightarrow U.$

Using the expression

$L^{-1}(\boldsymbol{\mathsf{p}}-u)=L\otimes(L^{\otimes-2}(3\boldsymbol{\mathsf{p}}))\otimes{\mathcal{O}}_{C}(-2\boldsymbol{\mathsf{p}}-u)=L\otimes{\mathcal{O}}_{C}(-2\boldsymbol{\mathsf{p}}-u+u+v+w)=L(v+w-2\boldsymbol{\mathsf{p}}),$

we are looking for a map

$\pi^{*}(L)(\tilde{v}+\tau\tilde{v}+\tilde{w}+\tau\tilde{w})\rightarrow\pi^{*}(L\otimes\omega_{C})\otimes{\mathcal{O}}_{{\widetilde{C}}}(\tilde{u}+\tilde{v}+\tilde{w}).$
(26)

Here, as usual, $\tau$ denotes the covering involution of the spectral cover
$\pi:{\widetilde{C}}\to C$.

Having a map (26) is equivalent to asking for a section of

$\pi^{*}(\omega_{C})\otimes{\mathcal{O}}_{{\widetilde{C}}}(\tilde{u}-\tau\tilde{v}-\tau\tilde{w}).$

The ramified cover $\pi:{\widetilde{C}}\rightarrow C$ has $4$ ramification
points $a,a^{\prime},b,b^{\prime}\in{\widetilde{C}}$ such that
$\pi(a),\pi(a^{\prime})$ and $\pi(b),\pi(b^{\prime})$ are opposite pairs under
the hyperelliptic involution of $C$.

Recall now that ${\widetilde{C}}$ itself is hyperelliptic of genus $5$, with
the map $\text{\sf
h}_{{\widetilde{C}}}:{\widetilde{C}}\rightarrow{\mathbb{P}}^{1}$ branched over
the $12$ inverse images in ${\widetilde{C}}$ of the $6$ Weierstrass points of
$C$. We have $\omega_{{\widetilde{C}}}=\text{\sf
h}_{{\widetilde{C}}}^{*}{\mathcal{O}}_{{\mathbb{P}}^{1}}(4)$. The
hyperelliptic involution $\sigma$ of ${\widetilde{C}}$ is one of the lifts of
the hyperelliptic involution of $C$, so $a,a^{\prime}$ and $b,b^{\prime}$ are
also opposite pairs under the hyperelliptic involution of ${\widetilde{C}}$.
It follows that
${\mathcal{O}}_{{\widetilde{C}}}(a+a^{\prime})\cong{\mathcal{O}}_{{\widetilde{C}}}(b+b^{\prime})\cong\text{\sf
h}_{{\widetilde{C}}}^{*}{\mathcal{O}}_{{\mathbb{P}}^{1}}(1)$. We get

$\pi^{*}(\omega_{C})\cong\omega_{{\widetilde{C}}}(-a-a^{\prime}-b-b^{\prime})=\text{\sf
h}_{{\widetilde{C}}}^{*}{\mathcal{O}}_{{\mathbb{P}}^{1}}(2).$

In particular, $\pi^{*}(\omega_{C})$ admits a section vanishing at the two
points $\tau\tilde{v},\tau\tilde{w}$. This gives the required section of
$\pi^{*}(\omega_{C})(\tilde{u}-\tau\tilde{v}-\tau\tilde{w})$.

This (together with the analogous arguments for $v$ and $w$) completes the
proof that the divisor $u+v+w$ is the fiber of the trigonal map over
$x\in\ell$. Thus, our line bundle $U$ corresponds to one of the choices of $8$
liftings as in the lemma.

Running this argument backwards shows that each of the $8$ liftings
corresponds to a choice of $U$. This gives the isomorphism claimed in the
lemma. ∎

Let $Y_{\ell}:=Y\times_{X}\ell$. It is a degree $8$ ramified cover of $\ell$
whose general fiber is described by the lemma. From this, we can describe the
ramification of $Y_{\ell}/\ell$:

######  Corollary 4.16.

Suppose $\ell$ is general. The branch locus of $Y_{\ell}/\ell$ consists two
disjoint groups of points: the $x\in\ell$ of the branch locus of $C/\ell$,
plus the points that are images of the four points $y\in C$ of the branch
locus of ${\widetilde{C}}/C$. The former class of points is the intersection
of $\ell$ with the wobbly locus. As $\ell$ is general, the second group
consists of four distinct points.

###### Proof.

The line $\ell$ gives a trigonal map $C\rightarrow{\mathbb{P}}^{1}$, and by
Lemma 4.15, the fiber of $Y_{\ell}$ over a point $x\in\ell$ consists of the
$8$ liftings to ${\widetilde{C}}$ of the three points $u,v,w$ in the trigonal
fiber over $x$.

When $x$ is a branch point of $C/\ell$, as we have seen in Proposition 4.14 it
means $x\in\ell\,\cap{\rm Wob}$, and two of the three points come together,
let’s say $u=v$. The monodromy around such a point interchanges $u$ and $v$.
We may assume that $u=v$ is general in $C$, in particular it isn’t a
ramification point of ${\widetilde{C}}/C$ so we may identify the two sheets of
${\widetilde{C}}$ labeled with $0,1$ near this point, similarly near $w$. With
these identifications we may write the $8$ points as $(a,b,c)$ where
$a,b,c\in\\{0,1\\}$. The monodromy sends $(a,b,c)$ to $(b,a,c)$. This has two
transpositions $(1,0,c)\leftrightarrow(0,1,c)$ for $c=0,1$.

Suppose $x$ is the image of one of the branch points $z\in C$ of
${\widetilde{C}}/C$. For $\ell$ general we may assume that $x$ isn’t wobbly.
The trigonal fiber is $u,v,w$ with, say, $u$ the branch point. As we move
around such a point, the set of choices of lifting of $u$ undergoes a
transposition; there are four such transpositions corresponding to the various
liftings of $v,w$.

The last statement, that was also mentioned in the previous paragraph, is that
two different branch points of ${\widetilde{C}}/C$ are not contained in the
same trigonal fiber. Indeed, the branch points consist of two general pairs of
opposite points under the hyperelliptic involution. If the trigonal map
identified two such points, they would have to be two opposite points, but
that would mean that the trigonal map identifies hyperelliptically opposite
points, hence that it factors through the hyperelliptic map, which isn’t the
case. ∎

######  Lemma 4.17.

Suppose $V$ is a vector bundle corresponding to a general point in $X$. Let
$A_{1},\ldots,A_{4}$ be the points in $\operatorname{Jac}^{0}(C)$
corresponding to the four lines in $X$ through $V$. Then the sum of these
points in $\operatorname{Jac}^{0}(C)$ is the origin of
$\operatorname{Jac}^{0}(C)$.

###### Proof.

Let $A=A_{1}$ be one of the line bundles, so we have

$0\rightarrow A\rightarrow V\rightarrow
A^{\vee}(\boldsymbol{\mathsf{p}})\rightarrow 0.$

As we have seen above, the other $A_{i}$ are of the form
$A_{i}=A^{\vee}(\boldsymbol{\mathsf{p}}-t_{i})$ where $t_{2}+t_{3}+t_{4}$ is a
divisor in the linear system $A^{\otimes-2}(3\boldsymbol{\mathsf{p}})$. This
yields the equation in $\operatorname{Jac}(C)$

$t_{2}+t_{3}+t_{4}=[A^{\otimes-2}(3\boldsymbol{\mathsf{p}})]=3\boldsymbol{\mathsf{p}}-2[A].$

Now, the sum of all the points is the sum in $\operatorname{Jac}(C)$

${\rm
sum}=[A]+[A^{\vee}(\boldsymbol{\mathsf{p}}-q_{2})]+[A^{\vee}(\boldsymbol{\mathsf{p}}-t_{3})]+[A^{\vee}(\boldsymbol{\mathsf{p}}-t_{4})]$
$=[A]+3[A^{\vee}]+3\boldsymbol{\mathsf{p}}-(t_{2}+t_{3}+t_{4})$
$=[A]-3[A]+3\boldsymbol{\mathsf{p}}-(3\boldsymbol{\mathsf{p}}-2[A])=0.$

∎

Given two of the four lines corresponding to line bundles $A_{1}$ and $A_{2}$,
we obtain a point $q\in C$ such that
$A_{2}=(A_{1})^{\vee}(\boldsymbol{\mathsf{p}}-q)$, in other words
$A_{1}\otimes A_{2}={\mathcal{O}}_{C}(\boldsymbol{\mathsf{p}}-q)$. In view of
the lemma, if $A_{3}$ and $A_{4}$ are the other two points, we have
$A_{1}\otimes A_{2}\otimes A_{3}\otimes A_{4}={\mathcal{O}}_{C}$ so
$A_{3}\otimes
A_{4}={\mathcal{O}}_{C}(q-\boldsymbol{\mathsf{p}})={\mathcal{O}}_{C}(\boldsymbol{\mathsf{p}}-q^{\prime})$.
Thus, the point of $C$ corresponding to $A_{3},A_{4}$ is $q^{\prime}$ the
image of $q$ by the hyperelliptic involution.

######  Remark 4.18.

In particular, we obtain the conclusion that the point $V\in X_{1}$ yields a
well-defined triple of points in ${\mathbb{P}}^{1}$. This gives a map
$X_{1}\rightarrow Sym^{3}({\mathbb{P}}^{1})\cong{\mathbb{P}}^{3}$ that will be
useful in Section 13.5.

###  4.6 Local structure of the spectral cover along the cusp locus

Let $R^{\rm mov}\subset Y$ be the movable locus of the ramification of the map
$Y\rightarrow X$.

Recall that $c_{1}(\Omega^{1}_{X})=2H$ and
$c_{1}(\Omega^{1}_{Y})=\boldsymbol{\mathit{E}}$ since $Y$ is obtained from the
abelian variety ${\mathcal{P}}$ by blowing up a curve of codimension $2$.
Comparing these, we conclude that the ramification locus has class
$\boldsymbol{\mathit{E}}+2\boldsymbol{\mathit{F}}$.

######  Lemma 4.19.

The order of ramification of $Y\rightarrow X$ along $\boldsymbol{\mathit{E}}$
is two (i.e. it has simple ramification at a general point of
$\boldsymbol{\mathit{E}}$).

###### Proof.

If we assume that the map has ramification of order $a$ along
$\boldsymbol{\mathit{E}}$, we get

$2\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}}=(a-1)\boldsymbol{\mathit{E}}+[R^{\rm
mov}].$

On the other hand, $R^{\rm mov}$ is effective, so
$2\boldsymbol{\mathit{F}}-(a-2)\boldsymbol{\mathit{E}}$ is effective. If $a>2$
this says that $2\boldsymbol{\mathit{F}}-\boldsymbol{\mathit{E}}$ is
effective. We note that $2\boldsymbol{\mathit{F}}-\boldsymbol{\mathit{E}}$ is
not zero, since $\boldsymbol{\mathit{E}}$ is clearly not ample whereas
$\boldsymbol{\mathit{F}}$ is ample (it is the pullback of an ample divisor by
a finite map).

Since $\boldsymbol{\mathit{F}}$ is ample, if
$2\boldsymbol{\mathit{F}}-\boldsymbol{\mathit{E}}$ is effective then
$(2\boldsymbol{\mathit{F}}-\boldsymbol{\mathit{E}})\cdot\boldsymbol{\mathit{F}}^{2}>0$.
However, from Proposition 4.8 saying $\boldsymbol{\mathit{F}}^{3}=32$ and
$\boldsymbol{\mathit{E}}\boldsymbol{\mathit{F}}^{2}=64$ it gives
$(2\boldsymbol{\mathit{F}}-\boldsymbol{\mathit{E}})\cdot\boldsymbol{\mathit{F}}^{2}=0$,
contradicting the effectivity. ∎

######  Proposition 4.20.

For generic choice of point in the Hitchin base, the movable ramification
locus $R^{\rm mov}$ intersects $\boldsymbol{\mathit{E}}$ in a union of two
constant sections of the fibration
$\boldsymbol{\mathit{E}}\rightarrow{\widehat{C}}$ and $64$ fibers. The
constant sections correspond to the two points in ${\mathbb{P}}^{1}$ whose
inverse images in $C$ make up the $4$ branch points of ${\widetilde{C}}/C$.

###### Proof.

From above, the full ramification locus has class
$\boldsymbol{\mathit{E}}+2\boldsymbol{\mathit{F}}$ and the fixed part has
multiplicity $1$ along $\boldsymbol{\mathit{E}}$. Thus, the class of $R^{\rm
mov}$ is $2\boldsymbol{\mathit{F}}$.

We use the description of Corollary 4.16. We are interested here in points
where the movable ramification meets the intersection with the wobbly locus.

We can describe more precisely, over points where $\ell$ meets the wobbly
locus, the fiber of $Y_{\ell}$. Suppose it is a general point of the wobbly
locus, so the degree $3$ divisor in $C$ over $x\in\ell$ has the form $2u+v$.
Suppose $u$ and $v$ are not in the branch locus of ${\widetilde{C}}/C$. To
describe the liftings to points of $Y_{\ell}$ we need to consider liftings of
nearby divisors into ${\widetilde{C}}$: the piece $2u$ splits into
$u_{1}+u_{2}$ and this has $4$ liftings. The monodromy action permutes two of
those liftings and leaves fixed the two other ones, depending on whether
$\tilde{u}_{1}$ and $\tilde{u}_{2}$ are in different or the same sheets of
${\widetilde{C}}/C$. Then each of those configurations becomes doubled
depending on the lifting of $v$. We obtain $2$ branch points and $4$
unbranched points of $Y_{\ell}/\ell$ over $x\in\ell$. This is as expected.

The points of $\boldsymbol{\mathit{E}}_{\ell}:=\boldsymbol{\mathit{E}}\cap
Y_{\ell}$ are the two branch points.

On the other hand, the points of the movable ramification locus correspond to
divisors $u_{1}+u_{2}+v$ such that one of those points is a branch point of
${\widetilde{C}}/C$. When that branch point is $v$ it does not correspond to a
point of $\boldsymbol{\mathit{E}}_{\ell}$. We would like to describe the cases
where $u$ is a branch point of ${\widetilde{C}}/C$.

Suppose given a divisor $2u+v$ in the linear system associated to
$A^{\otimes-2}(3\boldsymbol{\mathsf{p}})$. Then as we saw above, the other
line bundle of degree $0$ associated to $u$, that we’ll now denote
$L:=A^{\prime}$, is of the form

$L=A^{\vee}(\boldsymbol{\mathsf{p}}-u).$

Our bundle $V$ is an up-Hecke of $A\oplus L$ at the point $u$. We can write

$0\rightarrow L\rightarrow V\rightarrow
L^{\vee}(\boldsymbol{\mathsf{p}})\rightarrow 0$

and $A=L^{\vee}(\boldsymbol{\mathsf{p}}-u)$. The extension is the one that
vanishes when pulled back to $A$, or equivalently vanishing in the extension
group $H^{1}(L^{\otimes 2}(u-\boldsymbol{\mathsf{p}}))$.

We have

$L^{\otimes
2}(\boldsymbol{\mathsf{p}})=A^{\otimes-2}(3\boldsymbol{\mathsf{p}}-2u)$

but

$A^{\otimes-2}(3\boldsymbol{\mathsf{p}})\cong{\mathcal{O}}_{C}(2u+v)$

so

$L^{\otimes 2}(\boldsymbol{\mathsf{p}})={\mathcal{O}}_{C}(v).$

This is the equation saying that the line associated to $L$ is in the wobbly
locus. Furthermore, the set of extension classes corresponding to points of
this line is

$H^{1}(L^{\otimes
2}(-\boldsymbol{\mathsf{p}}))=H^{1}({\mathcal{O}}_{C}(v-2\boldsymbol{\mathsf{p}}))=H^{1}({\mathcal{O}}_{C}(-v^{\prime}))$

which maps isomorphically to $H^{1}({\mathcal{O}}_{C})$. Here $v^{\prime}$ is
the image of $v$ by the hyperelliptic involution.

An extension vanishes at the point $u$ if and only if it maps to zero in
$H^{1}({\mathcal{O}}_{C}(u))$, which is equivalent to saying that it
corresponds to the point in
${\mathbb{P}}^{1}={\mathbb{P}}H^{1}({\mathcal{O}}_{C})$ that is the image of
$u$ under the hyperelliptic double cover.

We conclude that under the birational map from the wobbly locus to
${\mathbb{P}}^{1}$, the image of a point corresponding to the degree $3$
divisor $2u+v$ with respect to the trigonal covering corresponding to the
original line $\ell$, it the image of $u$ in ${\mathbb{P}}^{1}$ corresponding
to the hyperelliptic map.

Now, we are interested in cases where $u$ coincides with one of the branch
points of ${\widetilde{C}}/C$. We know from Proposition 3.1 that there are $4$
branch points in $C$ that are the inverse images of two points in
${\mathbb{P}}^{1}$. Therefore, the points in question are ones where $u$ maps
to one of these two points in ${\mathbb{P}}^{1}$.

We conclude from this discussion that the points of $\boldsymbol{\mathit{E}}$
over the wobbly locus that map to one of these two points in
${\mathbb{P}}^{1}$ are in $R^{\rm mov}\cap\boldsymbol{\mathit{E}}$.

We have made some genericity assumptions in the above discussion. These hold
for the points we have just described, since the two points in
${\mathbb{P}}^{1}$ are general because we chose a general spectral cover
${\widetilde{C}}/C$. This says that $R^{\rm mov}\cap\boldsymbol{\mathit{E}}$
contains at least two constant sections of the fibration
$\boldsymbol{\mathit{E}}\rightarrow{\widehat{C}}$.

We now rely on the intersection number calculations to conclude. Indeed, we
have seen that $\boldsymbol{\mathit{F}}\cap\boldsymbol{\mathit{E}}$ has the
class of $\hat{\mathfrak{c}}+32\mathsf{fib}$ where $\hat{\mathfrak{c}}$ is a
constant section. This may be re-verified as follows: this class intersects
the fiber in one point, since the fiber maps to a line in $X$ that meets the
hyperplane class in one point. Our class is thus of the form
$\hat{\mathfrak{c}}+a\cdot\mathsf{fib}$. On the other hand, the self-
intersection of this class on
$\boldsymbol{\mathit{E}}\cong{\mathbb{P}}^{1}\times{\widehat{C}}$ is equal to
the intersection number $\boldsymbol{\mathit{F}}^{2}\boldsymbol{\mathit{E}}$
on $Y$; that is $64$. The self intersection of
$\hat{\mathfrak{c}}+a\cdot\mathsf{fib}$ is $2a$ giving $a=32$.

This shows that $2\boldsymbol{\mathit{F}}\cap\boldsymbol{\mathit{E}}$ is a
divisor in the class $2\hat{\mathfrak{c}}+64\mathsf{fib}$. Since we have
already exhibited a subset of this divisor consisting of two sections, we
conclude that it has to be of the form two sections plus $64$ fibers. This
completes the proof of the proposition. ∎

For a point $\mathsf{b}$ in the Hitchin base, let ${\rm Branch}^{\rm
mov}(\mathsf{b})\subset X$ be the image in $X$ of the ramification locus
$R^{\rm mov}\subset Y$ for the point $\mathsf{b}$. These form a family of
closed subvarieties of $X$.

We define the base locus of this family to consist of the set of points $x\in
X$ that lie in ${\rm Branch}^{\rm mov}(\mathsf{b})$ for $\mathsf{b}$ general
with respect to $x$, or equivalently for all $\mathsf{b}$.

######  Corollary 4.21.

If the base locus contains an irreducible subvariety of codimension $\leq 2$
then it is one of the lines in the wobbly locus.

###### Proof.

If $x\in X$ is not on the wobbly locus then it is a very stable bundle, and
the space of Higgs fields on that bundle maps properly to the Hitchin base.
Thus, the intersection with a general Hitchin fiber over $\mathsf{b}$ is
transverse, so $x$ is not in the branch locus of the map $Y\rightarrow X$ for
general $\mathsf{b}$.

Suppose $x$ is in the wobbly locus and let $S$ be the space of Higgs fields on
the corresponding bundle. It maps to the Hitchin base, by a map that is not
proper. However, for a general point $\mathsf{b}$ in the Hitchin base, the
intersection of the fiber over $\mathsf{b}$ with $S$ is transverse. Therefore,
for such a general $\mathsf{b}$, if $y\in Y$ is a point over $x$ that is not
in $\boldsymbol{\mathit{E}}$, then it is in
$S\cap\boldsymbol{h}^{-1}(\mathsf{b})$ hence not in $R^{\rm mov}$. This shows
that the points of $R^{\rm mov}$ lying over $x$ are in $R^{\rm
mov}\cap\boldsymbol{\mathit{E}}$.

We have seen in the proposition that for $\mathsf{b}$ general, the
intersection of $R^{\rm mov}$ with $\boldsymbol{\mathit{E}}$ consists of two
constant sections of the map
$\boldsymbol{\mathit{E}}={\widehat{C}}\times{\mathbb{P}}^{1}\to{\widehat{C}}$,
that move as a function of $\mathsf{b}$, and of $64$ fibers of the map
$\boldsymbol{\mathit{E}}={\widehat{C}}\times{\mathbb{P}}^{1}\to{\widehat{C}}$.
The images of these in the wobbly locus have the same description.

So, suppose $x$ is a point of the wobbly locus contained in the image of
$R^{\rm mov}$ for two different general values of $\mathsf{b}$. Then it is in
the intersection of two subsets of the form $2$ sections plus images of $64$
fibers; but the sections move, so $x$ has to be contained in at most one of
the $64$ lines. This completes the proof of the corollary. ∎

######  Corollary 4.22.

A general point on the cuspidal locus of the wobbly locus is not contained in
the movable part of the branch locus.

###### Proof.

This follows immediately from the previous corollary since the cuspidal locus
has codimension $2$ and is not a union of fibers. ∎

######  Corollary 4.23.

The “movable part” of the ramification locus does really move, in the sense
that the base locus does not contain any divisors.

###### Proof.

The above corollary implies that the codimension of the base locus is $\leq
2$. ∎

###  4.7 The step in the parabolic filtration

Recall that ${\mathcal{V}}_{a,b}=f_{*}({\mathcal{L}}_{a,b})$ and consider the
sheaf

${\mathcal{V}}_{a,b}/{\mathcal{V}}_{a,b-1}$

which is supported on ${\rm Wob}\subset X$ by definition. Over $X^{\circ}$
this is the quotient that will be used to define the crude parabolic
structure. For the computations, it will be necessary to have a refined
definition that holds over all of $X$. Recall that
$\boldsymbol{\mathsf{f}}:D\rightarrow X$ denotes the map from the
normalization $D$ of ${\rm Wob}$.

######  Lemma 4.24.

There is a quotient defined over $D$,

$\boldsymbol{\mathsf{f}}^{*}({\mathcal{V}}_{a,b})\rightarrow{\mathcal{U}}_{a,b}$

such that ${\mathcal{U}}_{a,b}$ is a rank $2$ torsion-free sheaf on $D$, hence
a bundle outside of a finite collection of points. Over $X^{\circ}$ (this
restriction being denoted by a superscript as usual) we have
$\left({\mathcal{V}}_{a,b}/{\mathcal{V}}_{a,b-1}\right)|_{X^{\circ}}=\boldsymbol{\mathsf{f}}_{*}{\mathcal{U}}^{\circ}_{a,b}$
as quotients of ${\mathcal{V}}^{\circ}_{a,b}$ and this characterizes
${\mathcal{U}}_{a,b}$.

###### Proof.

Start by noting that the property of the lemma uniquely characterizes
${\mathcal{U}}_{a,b}$. Indeed, the condition at the end states that the kernel
of the map
$\boldsymbol{\mathsf{f}}^{*}({\mathcal{V}}_{a,b})\rightarrow{\mathcal{U}}_{a,b}$
is, over $X^{\circ}$ (where $D^{\circ}={\rm Wob}^{\circ}\subset X$) the
subsheaf of sections generated by ${\mathcal{V}}_{a,b-1}$. The condition that
the quotient ${\mathcal{U}}_{a,b}$ is a torsion-free sheaf means that the
kernel is a saturated subsheaf, so it is defined by its restriction to
$D^{\circ}$.

This discussion tells us how to construct ${\mathcal{U}}_{a,b}$, namely set

${\mathcal{U}}^{\circ}_{a,b}:=\boldsymbol{\mathsf{f}}^{*}\left(\left({\mathcal{V}}_{a,b}/{\mathcal{V}}_{a,b-1}\right)|_{X^{\circ}}\right)=\left({\mathcal{V}}_{a,b}/{\mathcal{V}}_{a,b-1}\right)|_{X^{\circ}}$

and then let ${\mathcal{U}}_{a,b}$ be the unique torsion-free quotient of
$\boldsymbol{\mathsf{f}}^{*}{\mathcal{V}}_{a,b}$ that restricts to
${\mathcal{U}}^{\circ}_{a,b}$ over $D^{\circ}$. It is constructed by taking
the saturated subsheaf extending the kernel and taking the quotient.

For this discussion, one should note that ${\mathcal{V}}_{a,b}$ is a reflexive
sheaf on $X$, indeed it is the dirct image of a line bundle under a map from a
normal (and indeed smooth) variety to $X$. In particuler,
${\mathcal{V}}_{a,b}$ is a vector bundle outside of codimension $3$, therefore
$\boldsymbol{\mathsf{f}}^{*}({\mathcal{V}}_{a,b})$ is a vector bundle outside
of codimension $2$ on $D$.

The fact that ${\mathcal{U}}_{a,b}$ has rank $2$ will become apparent from the
next lemma. ∎

######  Lemma 4.25.

Let $f_{\boldsymbol{\mathit{E}}/D}:\boldsymbol{\mathit{E}}\rightarrow D$ be
the morphism induced by $f$. That is
$f_{|\boldsymbol{\mathit{E}}}:\boldsymbol{\mathit{E}}\to X$ factors as

$\textstyle{\boldsymbol{\mathit{E}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{\boldsymbol{\mathit{E}}/D}}$$\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\boldsymbol{\mathsf{f}}}$$\textstyle{X.}$

Then we have

${\mathcal{U}}_{a,b}=f_{\boldsymbol{\mathit{E}}/D,*}({\mathcal{L}}_{a,b}|_{\boldsymbol{\mathit{E}}}).$

###### Proof.

Over $X^{\circ}$, we have $D^{\circ}={\rm Wob}^{\circ}\subset X^{\circ}$ and
$\boldsymbol{\mathsf{f}}_{*}{\mathcal{U}}^{\circ}_{a,b}={\mathcal{U}}^{\circ}_{a,b}=f_{*}({\mathcal{L}}^{\circ}_{a,b})/f_{*}({\mathcal{L}}^{\circ}_{a,b}(-\boldsymbol{\mathit{E}}^{\circ}))$.
This quotient is equal, in turn, to
$f_{*}\left(i_{\boldsymbol{\mathit{E}}^{\circ},*}\left({\mathcal{L}}^{\circ}_{a,b}|_{\boldsymbol{\mathit{E}}^{\circ}}\right)\right)$,
where $i_{\boldsymbol{\mathit{E}}^{\circ}}:\boldsymbol{\mathit{E}}^{\circ}\to
Y$ is the natural inclusion. Over these smooth points
$\boldsymbol{\mathsf{f}}_{*}$ induces an equivalence between sheaves on
$D^{\circ}$ and sheaves on $X^{\circ}$ supported on $D^{\circ}$. Also,
$\boldsymbol{\mathsf{f}}\circ f_{\boldsymbol{\mathit{E}}/D}=f\circ
i_{\boldsymbol{\mathit{E}}}$ so

$\boldsymbol{\mathsf{f}}_{*}f_{\boldsymbol{\mathit{E}}^{\circ}/D^{\circ},*}({\mathcal{L}}^{\circ}_{a,b}|_{\boldsymbol{\mathit{E}}^{\circ}})=f_{*}(i_{\boldsymbol{\mathit{E}}^{\circ},*}{\mathcal{L}}^{\circ}_{a,b}|_{\boldsymbol{\mathit{E}}^{\circ}}).$

From the equivalence we conclude that

$f_{\boldsymbol{\mathit{E}}^{\circ}/D^{\circ},*}({\mathcal{L}}^{\circ}_{a,b}|_{\boldsymbol{\mathit{E}}^{\circ}})={\mathcal{U}}^{\circ}_{a,b}.$

This gives the claimed statement over $D^{\circ}$.

Next, consider the commutative diagram

$\textstyle{\boldsymbol{\mathit{E}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y\times_{X}D\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X}$

we see that ${\mathcal{V}}_{a,b}=f_{*}({\mathcal{L}}_{a,b})$ restricts over
$D$ to the direct image under the middle downward map, and this maps to
$f_{\boldsymbol{\mathit{E}}/D,*}({\mathcal{L}}_{a,b}|_{\boldsymbol{\mathit{E}}})$.
In other words, we have a map expressing the relationship in question. This
map is an isomorphism over $D^{\circ}$ by the preceding discussion. Since
$\boldsymbol{\mathit{E}}$ is smooth
$f_{\boldsymbol{\mathit{E}}/D,*}({\mathcal{L}}_{a,b}|_{\boldsymbol{\mathit{E}}})$
is torsion-free—in fact it is a bundle, by “miracle flatness”. Also, the map
$\boldsymbol{\mathit{E}}\rightarrow Y\times_{X}D$ is a closed embedding so

$f_{*}({\mathcal{L}}_{a,b})\rightarrow
f_{\boldsymbol{\mathit{E}}/D,*}({\mathcal{L}}_{a,b}|_{\boldsymbol{\mathit{E}}})$

is surjective. It follows that the right hand side is the unique torsion-free
quotient sheaf extending the quotient ${\mathcal{U}}^{\circ}_{a,b}$ already
known over $D^{\circ}$. This completes the proof that the right hand side is
${\mathcal{U}}_{a,b}$. Notice that we obtain ${\mathcal{U}}_{a,b}$ as a
torsion free sheaf of rank $2$ because $\boldsymbol{\mathit{E}}\rightarrow D$
is a $2:1$ covering. ∎

The next step is to use this characterization to calculate the Chern character
of ${\mathcal{U}}_{a,b}$. This calculation may be truncated at codimension $1$
on $D$ since that corresponds to codimension $2$ on $X$.

######  Proposition 4.26.

The Chern character of ${\mathcal{U}}_{a,b}$ truncated to codimension $1$ on
$D$ is

${\rm ch}({\mathcal{U}}_{a,b})=2+(2a+b)H_{D}-(3b+2)H_{D}^{\perp}$

where $H_{D}^{\perp}=[{\overline{C}}]-16[{\mathbb{P}}^{1}]$ is a class whose
direct image to $X$ vanishes.

###### Proof.

Write

$H_{D}:=\boldsymbol{\mathsf{f}}^{*}(H)=\alpha[{\overline{C}}]+\beta[{\mathbb{P}}^{1}].$

Noting that $H_{D}\cdot[{\mathbb{P}}^{1}]=1$ gives $\alpha=1$, then noting
that

$H_{D}\cdot H_{D}=H^{2}\cdot{\rm Wob}=32$

(see Lemma 4.4), we get $2\alpha\beta=32$ so $\beta=16$. Thus

$H_{D}=\boldsymbol{\mathsf{f}}^{*}(H)=[{\overline{C}}]+16[{\mathbb{P}}^{1}].$

We have

$\boldsymbol{\mathsf{f}}_{*}[{\overline{C}}]\cdot H=[{\overline{C}}]\cdot
H_{D}=16$

so (recalling $H^{3}=4$) we get
$\boldsymbol{\mathsf{f}}_{*}[{\overline{C}}]=4H^{2}$. We have that
$\boldsymbol{\mathsf{f}}_{*}[{\mathbb{P}}^{1}]$ is the class of a line in $X$,
this is $H^{2}/4$. We get

$\boldsymbol{\mathsf{f}}_{*}(H_{D})=8H^{2}.$

The class $H_{D}^{\perp}:=[{\overline{C}}]-16[{\mathbb{P}}^{1}]$ has the
property that $\boldsymbol{\mathsf{f}}_{*}(H_{D}^{\perp})=0$.

Turn to the question of the proposition. Consider the diagram

$\textstyle{\boldsymbol{\mathit{E}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i_{\boldsymbol{\mathit{E}}}}$$\scriptstyle{f_{\boldsymbol{\mathit{E}}/D}}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\boldsymbol{\mathsf{f}}}$$\textstyle{X.}$

We have

$i_{\boldsymbol{\mathit{E}},*}({\mathcal{L}}_{a,b}|_{\boldsymbol{\mathit{E}}})={\mathcal{L}}_{a,b}/{\mathcal{L}}_{a,b-1}.$

Applying $f_{*}$ to this gives ${\mathcal{V}}_{a,b}/{\mathcal{V}}_{a,b-1}$.

Recall that
${\mathcal{U}}_{a,b}:=f_{\boldsymbol{\mathit{E}}/D,*}({\mathcal{L}}_{a,b}|_{\boldsymbol{\mathit{E}}})$.
From the diagram,

$\boldsymbol{\mathsf{f}}_{*}({\mathcal{U}}_{a,b})\cong{\mathcal{V}}_{a,b}/{\mathcal{V}}_{a,b-1}.$

Therefore, in order to calculate ${\rm ch}({\mathcal{U}}_{a,b})$ we would like
to know the relationship between it and ${\rm
ch}(\boldsymbol{\mathsf{f}}_{*}({\mathcal{U}}_{a,b})$. Some further
information will be needed due to the fact that the codimension $1$ cycle
class group of $D\cong{\overline{C}}\times{\mathbb{P}}^{1}$ does not inject
into that of $X$.

First apply the Grothendieck-Riemann-Roch formula for
$\boldsymbol{\mathsf{f}}$. Recall that

${\rm td}(TX)^{-1}=1-H+5H^{2}/12.$

Truncated to codimension $1$ we get $\boldsymbol{\mathsf{f}}^{*}{\rm
td}(TX)^{-1}=1-H_{D}$. On the other hand, $2g_{{\overline{C}}}-2=32$ so
$c_{1}(TD)=2[{\overline{C}}]-32[{\mathbb{P}}^{1}]$, and

${\rm td}(TD)=1+[{\overline{C}}]-16[{\mathbb{P}}^{1}].$

The Grothendieck-Riemann-Roch formula says

$\displaystyle{\rm ch}(\boldsymbol{\mathsf{f}}_{*}({\mathcal{U}}_{a,b})$
$\displaystyle=\boldsymbol{\mathsf{f}}_{*}(2+c_{1}({\mathcal{U}}_{a,b}))(1+[{\overline{C}}]-16[{\mathbb{P}}^{1}])(1-H_{D})$
$\displaystyle=\boldsymbol{\mathsf{f}}_{*}\left(2+c_{1}({\mathcal{U}}_{a,b})+2[{\overline{C}}]-32[{\mathbb{P}}^{1}]-2[{\overline{C}}]-32[{\mathbb{P}}^{1}]\right).$

Noting that $\boldsymbol{\mathsf{f}}_{*}(2[D])=2[{\rm Wob}]$ and recalling
from Lemma 4.4 that $[{\rm Wob}]=8H$, then also using
$\boldsymbol{\mathsf{f}}_{*}[{\mathbb{P}}^{1}]=H^{2}/4$, we get

${\rm
ch}(\boldsymbol{\mathsf{f}}_{*}({\mathcal{U}}_{a,b})=16H+i_{*}c_{1}({\mathcal{U}}_{a,b})-16H^{2}.$

Denote by $\Delta_{b}$ the operation on a function $f$,
$\Delta_{b}(f):=f(a,b)-f(a,b-1)$. Thus for example $\Delta_{b}(b)=1$ and
$\Delta_{b}(b^{2})=2b-1$ With this notation,

$\displaystyle{\rm ch}({\mathcal{V}}_{a,b}/{\mathcal{V}}_{a,b-1})$
$\displaystyle=\Delta_{b}\left[8+(8a+16b)H+(4a^{2}+16ab+4b^{2}-12b-2)H^{2}\right]$
$\displaystyle=16H+(16a+8b-16)H^{2}.$

The comparison ${\rm ch}(i_{*}({\mathcal{U}}_{a,b})={\rm
ch}({\mathcal{V}}_{a,b}/{\mathcal{V}}_{a,b-1})$ yields

$\boldsymbol{\mathsf{f}}_{*}c_{1}({\mathcal{U}}_{a,b})=(16a+8b)H^{2}.$

In particular, it follows that

$c_{1}({\mathcal{U}}_{a,b})=(2a+b)H_{D}+\xi H_{D}^{\perp}$

for some $\xi$.

On the other hand, we can note that the inverse image in
$\boldsymbol{\mathit{E}}$ of a line ${\mathbb{P}}^{1}\subset D$ consists of
two fibers in $\boldsymbol{\mathit{E}}$. A fiber maps to a line in $X$ so its
intersection with $\boldsymbol{\mathit{F}}$ has one point. On the other hand,
$\boldsymbol{\mathit{E}}$ intersected with a fiber is $-1$. Thus, the bundle
${\mathcal{L}}_{a,b}$ restricted to a fiber has class ${\mathcal{O}}_{\rm
fib}(a-b-1)$ so its direct image to the line ${\mathbb{P}}^{1}\subset D$ has
$c_{1}=2c_{1}[{\mathcal{O}}_{{\mathbb{P}}^{1}}(a-b-1)]$. This tells us that

$c_{1}({\mathcal{U}}_{a,b}|_{{\mathbb{P}}^{1}})=2a-2b-2.$

Thus

$((2a+b)H_{D}+\xi H_{D}^{\perp}).[{\mathbb{P}}^{1}]=2a-2b-2.$

This gives

$(2a+b)+\xi=2a-2b-2$

so $\xi=-2-3b$ and we get

$c_{1}({\mathcal{U}}_{a,b})=(2a+b)H_{D}-(3b+2)H_{D}^{\perp}.$

∎

From this calculation we immediately get the following

######  Corollary 4.27.

Suppose $S\subset D$ is a curve in the class of $kD_{H}$, and take $2a+b=0$.
Then ${\mathcal{U}}_{a,b}|_{S}$ is a rank $2$ vector bundle of degree $0$ on
$S$.

###  4.8 Hyperplane section

Let us now intersect the wobbly locus ${\rm Wob}\subset X$ with a general
hyperplane section $X_{H}$, this denoting a smooth divisor in the class of $H$
on $X$.

The class of the divisor $H$ on $D$ is $H_{D}$, described above as

$H_{D}=[{\overline{C}}]+16[{\mathbb{P}}^{1}].$

One has $(H_{D})^{2}=32$. We note also that $H_{D}$ is the trace of $X_{H}$ on
$D$, in particular its image in $X_{H}$ is ${\rm Wob}\cap X_{H}$. That is a
curve ${\rm Wob}_{H}$ in $X_{H}$ whose normalization is isomorphic to
${\overline{C}}$ since the projection $H_{D}\rightarrow{\overline{C}}$ is an
isomorphism.

We would now like to count the double points of the curve ${\rm Wob}_{H}$. The
intersection of the curve with the $6$ divisors

${\overline{C}}\times\\{\boldsymbol{\mathsf{p}}_{i}\\}$

are the points in the normalization ${\overline{C}}$ that get identified in
pairs to form the double points of the image curve ${\rm Wob}_{H}={\rm
Wob}\cap X_{H}$. The number of points in the intersection of $H_{D}$ with
these $6$ horizontal divisors is $6\cdot 16=96$, so there are $96$ points
lying over the double points. We conclude:

  * •

There are $48$ double points on the curve ${\rm Wob}_{H}$.

We now consider the genus of the normalization of ${\rm Wob}_{H}$. The class
of ${\rm Wob}_{H}$ is in $8H$ on $X_{H}$, so

$({\rm Wob}_{H})^{2}=(8H)^{2}_{X_{H}}=64H^{3}=64\cdot 4=256.$

The genus of a smooth curve $S$ in this class is given as follows: we recall
that $K_{X_{H}}=-H$ so

$K_{S}=K_{X_{H}}\cdot
S+{\mathcal{O}}_{S}(S)=-H\cdot(8H)_{X_{H}}+(8H)\cdot(8H)_{X_{H}}=56H^{3}=224.$

This gives $2g_{S}-2=224$ so $g_{S}=113$.

If there were no singularities except the $48$ double points, we would get the
genus of the normalization of ${\rm Wob}_{H}$ to be $113-48=65$. However, the
normalization is ${\overline{C}}$ that has genus $17$. Hence, there are
additional singularities accounting for a drop in the genus of $65-17=48$. We
first guessed then proved:

  * •

The curve ${\rm Wob}_{H}$ has its $48$ nodes, plus $48$ cusps.

The count of cusps comes from the structural result on the cuspidal locus in
Proposition 4.5, proven in Subsection 4.4. The number of cusps of is obtained
as follows. The structural result says that the cuspidal locus of ${\rm Wob}$
is the graph of a map ${\overline{C}}\rightarrow{\mathbb{P}}^{1}$ of degree
$32$. It intersects $[{\mathbb{P}}^{1}]$ in $1$ point and $[{\overline{C}}]$
in $32$ points so it intersects $H_{D}=[{\overline{C}}]+16[{\mathbb{P}}^{1}]$
in $48$ points.

###  4.9 Summary of properties of our given sheaves

We are defining a vector bundle ${\mathcal{V}}_{a,b}$ on $X$ by taking the
direct image of the line bundle
${\mathcal{L}}_{0}(a\boldsymbol{\mathit{F}}+(b+1)\boldsymbol{\mathit{E}})$ for
a line bundle ${\mathcal{L}}_{0}$ of degree $0$. We have calculated its Chern
character in Proposition 4.12,

${\rm ch}({\mathcal{V}}_{a,b})=8+(8a+16b)H+(4a^{2}+16ab+4b^{2}-12b-2)H^{2}.$

It is instructive to calculate the Bogomolov $\Delta$-invariant. Notice that
modifying $a$ corresponds to tensoring with a line bundle pulled back from
$X$, so we may assume $a=0$. Then

$\displaystyle\Delta({\mathcal{V}}_{0,b})$
$\displaystyle=\frac{1}{2r}c_{1}^{2}-{\rm ch}_{2}$
$\displaystyle=(16b^{2}-(4b^{2}-12b-2))H^{2}$
$\displaystyle=(12b^{2}+12b+2)H^{2}.$

For integer values of $b$, the minimum of $2H^{2}$ is attained at both values
$b=-1$ and $b=0$. Symmetry considerations suggest that the desired parabolic
weight should be $1/2$. with the parabolic structure creating an interpolation
from ${\mathcal{V}}_{0,-1}$ to ${\mathcal{V}}_{0,0}$.

One can see, by the way, that it is not correct to just imagine a fractional
value of $b$, indeed the above formula at $b=-1/2$ would give $\Delta=-H^{2}$
contradicting the Bogomolov-Gieseker inequality. Nonetheless, the symmetry
consideration leads us to try using a parabolic weight of $1/2$ which will
turn out to work, once the second Chern class is adjusted appropriately at the
singularities.

Over $D$ we have a morphism

$\boldsymbol{\mathsf{f}}^{*}({\mathcal{V}}_{a,b})\rightarrow{\mathcal{U}}_{a,b}\rightarrow
0$

and the Chern character of ${\mathcal{U}}_{a,b}$ on $D$ is

${\rm ch}({\mathcal{U}}_{a,b})=2+(2a+b)H_{D}-(3b+2)H_{D}^{\perp}$

by Proposition 4.26.

We would like to use this to define a parabolic structure along ${\rm Wob}$.
It will be easier now to pass to the hyperplane section $X_{H}$ and consider
the parabolic structure along ${\rm Wob}_{H}$. Let $D_{H}$ denote the
normalization of ${\rm Wob}_{H}$, and $H_{D_{H}}$ the restriction of $H$ to
here.

Note that $H_{D}^{\perp}$ restricts to zero on the hyperplane section $D_{H}$,
so

${\rm ch}({\mathcal{U}}_{a,b})|_{D_{H}}=2+(2a+b)H_{D_{H}}.$

It is not at all easy to understand how to calculate the contributions of
cusps to the parabolic $c_{2}$. Taher’s papers [Tah10, Tah13] consider a
multiple point, rather than a cusp, and in any case the procedure—which could
theoretically be implemented also for the cusp, or even imported by going to a
covering and transforming the cusp to a triple point—is rather complicated.

Therefore, we shall adopt a shortcut: we will just look at the calculation for
the parabolic weight $\alpha=1/2$, and we will do that by going to a cover
that is ramified with ramification index of $2$ over the curve ${\rm
Wob}_{H}$. Then, on the cover rather than defining a parabolic structure we
will use the quotient ${\mathcal{U}}$ to make a modification of the bundle.
The construction follows the technique described in Proposition 3.15.

###  4.10 First parabolic Chern class

The first parabolic Chern class may be described by a simplified integral
formula as just the average of ${\rm ch}_{1}({\mathcal{E}}_{a})$ over
$a\in[0,1]$. In our situation of a crude parabolic structure with sheaves
${\mathcal{V}}={\mathcal{V}}_{a,b}$ and
${\mathcal{V}}^{\prime}={\mathcal{V}}_{a,b-1}$ and parabolic weight
$\alpha=1/2$ the formula becomes

$\displaystyle{\rm ch}_{1}^{\rm par}({\mathcal{E}}_{\cdot})$
$\displaystyle=\int_{a=0}^{1}{\rm ch}_{1}({\mathcal{E}}_{a})da$
$\displaystyle=\frac{{\rm ch}_{1}({\mathcal{V}}^{\prime})+{\rm
ch}_{1}({\mathcal{V}})}{2}$ $\displaystyle=\frac{(8a+16(b-1))H+(8a+16b)H}{2}$
$\displaystyle=(8a+16b-8)H.$

We should choose $a=1$ and $b=0$ to give ${\rm ch}_{1}^{\rm par}=0$. This is
motivated by the symmetry consideration described above and the fact that we
are looking for a parabolic structure with vanishing first Chern class.

However, recall that modifying $a$ amounts to tensoring with a line bundle on
$X=X_{1}$. We can do that later. For simplicity we will therefore assume
$a=0,b=0$.

###  4.11 Going to a cover of $X_{H}$

In order to put this strategy into play, let us suppose we are given a finite
ramified covering $g:Z\rightarrow X_{H}$ together with a ramification locus
$R\subset Z$, such that $g^{-1}({\rm Wob}_{H})=2R$, in other words $g$ is
ramified with ramification of order $2$ along $R$ over ${\rm Wob}_{H}$.

We recall that the cusp is the downstairs branch locus of the quotient

${\mathbb{C}}^{2}\rightarrow{\mathbb{C}}^{2}/S_{3}={\mathbb{C}}^{2}.$

Upstairs, the branch locus consists of $3$ crossed lines. Therefore, we may
assume that this provides the local model for our covering $Z\rightarrow
X_{H}$ over cusps of ${\rm Wob}_{H}$. This local model is a covering of degree
$6$.

On the other hand, we will suppose that the local model for the covering over
a double point, is just the product of two ramified coverings with
ramification two. This local model is a covering of degree $4$.

The covering $Z$ may be constructed globally by the Kawamata covering trick
[Kaw88]. There may be ramification elsewhere but that does not need to worry
us.

Let $d$ denote the degree of the covering $Z\to X_{H}$.

  * •

Over the $48$ cusps of ${\rm Wob}_{H}$ we will therefore have $48d/6=8d$
triple points of $R$.

  * •

Over the $48$ double points of ${\rm Wob}_{H}$ we will in turn have
$48d/4=12d$ double points of $R$.

Let $\widetilde{R}$ denote the normalization of $R$. Let ${\mathcal{V}}_{Z}$
denote the restriction of ${\mathcal{V}}={\mathcal{V}}_{a,b}$ to $Z$. We have
a map, defined over $\widetilde{R}$:

${\mathcal{V}}_{Z}|_{\widetilde{R}}\rightarrow{\mathcal{U}}_{\widetilde{R}}.$

Here ${\mathcal{U}}_{\widetilde{R}}$ is the pullback of ${\mathcal{U}}_{a,b}$
to $\widetilde{R}$. Let

$i_{\widetilde{R}}:\widetilde{R}\rightarrow Z$

denote the birational inclusion. We can define the “normal bundle”
$N_{\widetilde{R}/Z}$ by

$i_{\widetilde{R}}^{*}K_{Z}\otimes N_{\widetilde{R}/Z}\cong
K_{\widetilde{R}}.$

Let us calculate this in the following way. From the picture, a double point
leads to two extra self-intersection points, whereas a triple point leads to
$6$ extra self-intersection points. Therefore, we get

$N_{\widetilde{B}/Z}+2(\mbox{double pts})+6(\mbox{triple pts})=R^{2}.$

Now $R$ has class $4H$ so $R^{2}=(4H)^{2}=16H^{2}$, which pulled back to $Z$
together with $H^{2}=4$ on $X_{H}$, gives

$R^{2}=64d.$

Recall that we have $12d$ double points and $8d$ triple points, so our formula
gives

$\displaystyle\deg N_{\widetilde{R}/Z}$ $\displaystyle=64d-2\cdot 12d-6\cdot
8d$ $\displaystyle=64d-24d-48d$ $\displaystyle=-8d.$

On the other hand, let us recall the Grothendieck-Riemann-Roch formula
(ignoring higher order terms)

${\rm
ch}(i_{\widetilde{R}*}{\mathcal{U}}_{\widetilde{R}})=i_{\widetilde{R}*}{\rm
ch}({\mathcal{U}}_{\widetilde{R}})-r_{{\mathcal{U}}_{\widetilde{R}}}i_{\widetilde{R}*}N_{\widetilde{R}/Z}/2.$

Recall that ${\mathcal{U}}$ has rank $2$. Thus the term
$-r_{{\mathcal{U}}_{\widetilde{R}}}i_{\widetilde{R}*}N_{\widetilde{R}/Z}/2$ is
equal to $-(-8d)=+8d$ and so we get

${\rm
ch}(i_{\widetilde{R}*}{\mathcal{U}}_{\widetilde{R}})=2R+\deg{\mathcal{U}}_{\widetilde{R}}+8d.$

Recall that $2R=g^{*}({\rm Wob}_{H})$ has class $8H$.

Let us now set $(a,b)=(0,0)$. As explained in the previous subsections, this
differs from the value we guess that it will be good to look at, by a
modification of $a$ that amounts to tensoring with a line bundle on $X$, and
we will recover that later.

With this choice, ${\mathcal{U}}$ has degree $0$ by Corollary 4.27, so we get

${\rm ch}(i_{\widetilde{R}*}{\mathcal{U}}_{\widetilde{R}})=8H+8d.$

We now note that the map

$V_{Z}\rightarrow i_{\widetilde{R}*}{\mathcal{U}}_{\widetilde{R}}$

is not going to be surjective.

Here is a guess as to what the image is going to be. The proof will be a
consequence of the Bogomolov-Gieseker inequality.

A little like the principle that the two quotients are independent at the
double points, it looks likely that the quotient over the double point is
something that is “moving”, corresponding to a smooth curve in the Grassmanian
of the total bundle above the cusp. The guess is that the image of the above
map corresponds then, not to the structure sheaf of the embedded plane curve
$R$, but rather to the version of $R$ that has a triple point embedded as a
space curve. Let us call this curve $R^{\sharp}$, it is obtained by not
glueing together the points lying over double points, but by glueing together
the points lying over a triple point in as small a way as possible. We get

$0\rightarrow{\mathcal{O}}_{R^{\sharp}}\rightarrow{\mathcal{O}}_{\widetilde{R}}\rightarrow{\mathcal{S}}\rightarrow
0$

where ${\mathcal{S}}$ is a skyscraper sheaf that has length $2$ over each
triple point. (To get to the plane curve $R$ one would use a skyscraper sheaf
of length $3$).

######  Lemma 4.28.

The image of the map

$V_{Z}\rightarrow i_{\widetilde{R}*}{\mathcal{U}}_{\widetilde{R}}$

is the subsheaf

${\mathcal{U}}_{R^{\sharp}}\subset
i_{\widetilde{R}*}{\mathcal{U}}_{\widetilde{R}}.$

The proof will be deferred to below, since it is motivated by the following
computations.

With this hypothesis, we have

$0\rightarrow{\mathcal{U}}_{R^{\sharp}}\rightarrow
i_{\widetilde{R}*}{\mathcal{U}}_{\widetilde{R}}\rightarrow{\mathcal{S}}_{U}\rightarrow
0$

where ${\mathcal{S}}_{U}$ has length $4$ over each triple point ($4$ because
$U$ has rank $2$).

Then, the hypothesis is that we have a surjection

$V_{Z}\rightarrow{\mathcal{U}}_{R^{\sharp}}\rightarrow 0.$

Let $V^{\prime}_{Z}$ be the kernel, this will be our vector bundle over the
covering $Z$ corresponding to the “pullback of the parabolic bundle to $Z$”.

Using the fact that there are $8d$ double points, we get

$\displaystyle{\rm ch}({\mathcal{U}}_{R^{\sharp}})$ $\displaystyle={\rm
ch}(i_{\widetilde{R}*}{\mathcal{U}}_{\widetilde{R}})-8d\cdot 4$
$\displaystyle=8H+8d-32d=8H-24d.$

Let us put this together with the Chern character of $V_{Z}$. Recall that we
are setting $(a,b)=(0,0)$ so ${\rm ch}_{2}({\mathcal{V}}_{0,0})=-2H^{2}$ on
$X_{H}$. Recall that $H^{2}$ has $4$ points on $X_{H}$, and then we pullback
by the covering $g$ of degree $d$, so we get

${\rm ch}_{2}(V_{Z})=-8d$

(here measuring in terms of numbers of points) on $Z$.

Recall also that at $(a,b)=(0,0)$ we have ${\rm
ch}_{1}({\mathcal{V}}_{0,0})=0$.

Altogether:

$\displaystyle{\rm ch}(V^{\prime}_{Z})$ $\displaystyle={\rm ch}(V_{Z})-{\rm
ch}({\mathcal{U}}_{R^{\sharp}})$ $\displaystyle=8-8H-8d+24d$
$\displaystyle=8-8H+16d$ $\displaystyle=8(1-H+2d).$

We now note that $H^{2}/2=2d$ points on $Z$ so, the term $2d$ above that is
counted in terms of points on $Z$, may be written as $H^{2}/2$. We get the
formula:

${\rm ch}(V^{\prime}_{Z})=8(1-H+H^{2}/2).$

This clearly expresses the bundle $V^{\prime}_{Z}$ as equivalent to $8$ times
a line bundle since $1-H+H^{2}/2=e^{-H}$. Therefore, just by inspection, its
$\Delta$-invariant vanishes and we have a projectively flat bundle.

If we then tensor with ${\mathcal{O}}_{X}(H)$ we obtain a Higgs bundle with
vanishing Chern classes.

####  4.11.1 Proof of the lemma

Following through what would happen in the above calculations if the image
were different, we can now give the proof of the lemma.

###### Proof of Lemma 4.28.

We note that the map $V_{Z}\rightarrow
i_{\widetilde{R}*}{\mathcal{U}}_{\widetilde{R}}$ sends sections of $V_{Z}$ to
sections of ${\mathcal{U}}_{\widetilde{R}}$ that agree over the point where
the three branches come together. Thus, the morphism factors through a map
$V_{Z}\rightarrow{\mathcal{U}}_{R^{\sharp}}$. The question is to show that it
is surjective. Suppose not. Let ${\mathcal{U}}^{\prime}$ denote the image, and
let $\ell$ denote the total length of the quotient sheaf
${\mathcal{U}}^{\prime}/{\mathcal{U}}_{R^{\sharp}}$ (take the sum over all the
cuspidal points).

In the above discussion, let $V^{\prime}_{Z}$ still denote the kernel of the
map $V_{Z}\rightarrow{\mathcal{U}}^{\prime}$, so we have a left exact sequence

$0\rightarrow V^{\prime}_{Z}\rightarrow
V_{Z}\rightarrow{\mathcal{U}}^{\prime}\rightarrow 0.$

Now ${\rm ch}({\mathcal{U}}^{\prime})={\rm
ch}({\mathcal{U}}_{R^{\sharp}})-\ell=8H-24d-\ell$, and following the same
calculation as before but with the extra term $\ell$ gives

$\displaystyle{\rm ch}(V^{\prime}_{Z})$ $\displaystyle={\rm ch}(V_{Z})-{\rm
ch}({\mathcal{U}}^{\prime})$ $\displaystyle=8-8H-8d+24d+\ell$
$\displaystyle=8(1-H+2d)+\ell$ $\displaystyle=8(1-H+H^{2}/2)+\ell.$

Recall that the Bogomolov-Gieseker inequality says that $c_{2}$ can only go up
from the flat case, hence ${\rm ch}_{2}$ can only go down. The bundle
$V^{\prime}_{Z}$ would be a stable Higgs bundle on $Z$, and if $\ell>0$ this
would contradict Bogomolov-Gieseker. Thus, we conclude $\ell=0$ which
completes the proof of the lemma. ∎

###  4.12 Degree one case—conclusion

######  Theorem 4.29.

Suppose ${\mathcal{L}}_{0}\in\operatorname{Pic}(Y)$ is the pullback of a flat
line bundle on the abelian variety ${\mathcal{P}}$, and set
${\mathcal{L}}:={\mathcal{L}}_{0}\otimes{\mathcal{O}}_{Y}(\boldsymbol{\mathit{E}}+\boldsymbol{\mathit{F}})$.
Let

${\mathcal{V}}:=f_{*}({\mathcal{L}})={\mathcal{V}}_{0,0}\otimes{\mathcal{O}}_{X}(H)$

provided with its natural crude parabolic structure with weight $\alpha=1/2$
(tensoring the one defined above with quotient ${\mathcal{U}}_{0,0}$, by
${\mathcal{O}}_{X}(H)$). Then this is a crude parabolic logarithmic Higgs
bundle over $(X^{\circ},{\rm Wob}^{\circ})$ that admits an extension to a
purely imaginary twistor ${\mathscr{D}}$-module as in Theorem 3.13. The
associated local system on $X-{\rm Wob}$ has rank $8$ and its monodromy around
${\rm Wob}$ is semisimple with eigenvalues $1$ of multiplicity $6$ and $-1$ of
multiplicity $2$.

The spectral line bundle on $Y$ is
${\mathcal{L}}^{\prime}={\mathcal{L}}_{0}\otimes{\mathcal{O}}_{Y}(\boldsymbol{\mathit{F}})$.

###### Proof.

We saw in the previous subsections that choosing ${\mathcal{V}}_{a,b}$ for
$(a,b)=(0,0)$ yields a projectively flat solution whose (truncated) parabolic
Chern character is $8(1-H+H^{2}/2)$.

Tensoring this by ${\mathcal{O}}_{X_{1}}(H)$ corresponds to choosing
$(a,b)=(1,0)$, taking the crude parabolic structure ${\mathcal{V}}_{1,0}$ with
subsheaf ${\mathcal{V}}^{\prime}_{1,0}={\mathcal{V}}_{1,-1}$, and this yields
a parabolic Higgs bundle with vanishing parabolic Chern classes. The rank is
$8$ and the parabolic structure along ${\rm Wob}$ has parabolic weight $1/2$
with multiplicity equal to the rank of ${\mathcal{U}}$ which is $2$, completed
by parabolic weight $0$ with multiplicity $6$. This gives the stated
eigenvalues of the monodromy.

We note that the spectral line bundle, by definition the one whose direct
image from $Y$ to $X$ gives ${\mathcal{E}}_{0}$, is
${\mathcal{L}}^{\prime}_{a,b}={\mathcal{L}}_{a,b-1}={\mathcal{L}}_{0}\otimes{\mathcal{O}}_{Y}(\boldsymbol{\mathit{F}})$.
∎

##  5 The degree zero moduli space

In this section $X$ denotes the degree $0$ moduli space and $Y$ denotes its
modular spectral covering of degree $8$ that is the blow-up of the Prym
variety ${\mathcal{P}}={\mathcal{P}}_{2}$ in $16$ points. Similarly ${\rm
Wob}$ means ${\rm Wob}_{0}$ and so forth.

Narasimhan and Ramanan show that $X\cong{\mathbb{P}}^{3}$ [NR69]. There is a
_Kummer surface_ of degree $4$ with $16$ nodes that we will denote by
$\mathsf{Kum}\subset X$. This is the moduli space of bundles that are
semistable but not stable, up to $S$-equivalence. The $S$-equivalence class
corresponding to a point of $\mathsf{Kum}$ is represented by a polystable
bundle of the form $L\oplus L^{\vee}$ where $L\in\operatorname{Jac}^{0}(C)$ is
a line bundle of degree $0$.

We have

$\mathsf{Kum}\cong\operatorname{Jac}^{0}(C)/\pm 1,$

a point on $\mathsf{Kum}$ corresponding to a pair $(L,L^{-1})$ of degree zero
line bundles on $X$, up to interchangeing $L$ and $L^{-1}$. The corresponding
polystable bundle is $E=L\oplus L^{-1}$. When $L=L^{-1}$, that is to say at
the $16$ points of order two on $\operatorname{Jac}^{0}(C)$, we get a singular
point of $\mathsf{Kum}$.

Let $\mathsf{Higgs}_{0}$ be the corresponding Hitchin moduli space of Higgs
bundles with determinant ${\mathcal{O}}_{C}$, the Higgs field having zero
trace.

###  5.1 Geometry of the wobbly locus in degree zero

The Higgs bundle moduli space has a locus analogous to the Kummer surface.

######  Lemma 5.1.

The Hitchin moduli space is singular along a codimension two subvariety
$\mathsf{Higgs}_{0}^{\rm sing}$, and this subvariety consists of the set of
polystable but not stable Higgs bundles. We have

$\mathsf{Higgs}_{0}^{\rm sing}\cong\frac{\operatorname{Jac}^{0}(C)\times
H^{0}(C,K_{C})}{\pm 1},$

a point here parametrizing a Higgs bundle of the form
$(L,\phi)\oplus(L^{-1},-\phi)$. The set of ${\mathbb{C}}^{\times}$ fixed
points of $\mathsf{Higgs}_{0}^{\rm sing}$ is equal to the Kummer surface
$\mathsf{Kum}\subset X$. Outside of the $16$ singular points of
$\mathsf{Kum}$, the transverse local structure of $\mathsf{Higgs}_{0}$ along
$\mathsf{Higgs}_{0}^{\rm sing}$ is that of a simple double point.

###### Proof.

A stable Higgs bundle $\mathbb{E}$ has only scalar endomorphisms, and by
duality the trace-free $H^{2}$, that is to say $H^{2}_{\rm
Dol}(End_{0}(\mathbb{E}))$, vanishes. This is the obstruction space to
deforming $\mathbb{E}$. Thus, a stable Higgs bundle is a smooth point of
$\mathsf{Higgs}_{0}$ (this is well-known, of course). Thus, a singular point
must be a strictly semistable point represented by a polystable Higgs bundle.
Since we are in rank $2$, it is a direct sum of line bundles, and the
condition that the determinant is trivial means that it must have the stated
form $(L,\phi)\oplus(L^{-1},-\phi)$. The structure is that of the moduli space
of $(L,\phi)$ modulo the involution $(L,\phi)\mapsto(L^{-1},-\phi)$.

A Higgs bundle of this form is fixed by ${\mathbb{C}}^{\times}$ if and only if
$\phi=0$, the fixed points are therefore inside the moduli space of bundles
$X\subset\mathsf{Higgs}_{0}$ and are in the Kummer surface.

Suppose we are at a point $\mathbb{L}\in\mathsf{Kum}$ corresponding to
$(L,\phi=0)$ with $L\not\cong L^{-1}$. In other words
$\mathbb{L}=(L,0)\oplus(L^{-1},0)$, $H^{2}_{\rm
Dol}(End^{0}(\mathbb{L}))={\mathbb{C}}$, and

$H^{1}_{\rm Dol}(End_{0}(\mathbb{L}))=H^{1}_{{\rm Dol}}({\mathcal{O}})\oplus
H^{1}_{{\rm Dol}}(L^{\otimes 2})\oplus H^{1}_{{\rm Dol}}(L^{\otimes-2})$

with the Kuranishi map being given by the trace of the cup-product (recall
that the Goldman-Millson deformation theory is formal [GM88, Sim92]). The cup
product vanishes on $H^{1}_{{\rm Dol}}({\mathcal{O}})$, which is the
unobstructed deformation space of $(L,0)$. On $H^{1}_{{\rm Dol}}(L^{\otimes
2})\oplus H^{1}_{{\mathsf{Dol}^{\operatorname{par}}_{L^{2}}}}(L^{\otimes-2})$
it is the same as the Poincaré duality form. Thus, as a quadratic form it
defines a simple double point in the direction transverse to the deformation
space of $(L,0)$ i.e. the tangent space of $\mathsf{Higgs}_{0}^{\rm sing}$. We
see in particular that $\mathsf{Higgs}_{0}$ is indeed singular along this
locus. ∎

In Proposition 3.3(a) we showed that the set
$\mathsf{Higgs}_{0}^{{\mathbb{C}}^{\times},\text{nu}}$ of
${\mathbb{C}}^{\times}$-fixed points on $\mathsf{Higgs}_{0}$ that are not on
$X$, is a disjoint union of $16$ points. These points cprrespond to the $16$
Higgs bundles

$\mathbb{E}_{\kappa}=(E_{\kappa},\theta_{\kappa})=\left(\kappa\oplus\kappa^{-1},\,\begin{pmatrix}0&0\\\
1&0\end{pmatrix}\right),\quad\kappa\in\mathsf{Spin}(C).$

From this picture, we can now describe the wobbly locus in $X$ completely.

######  Proposition 5.2.

The wobbly locus ${\rm Wob}\subset X$ decomposes as

${\rm
Wob}=\mathsf{Kum}\cup\left(\bigcup_{\kappa\in\mathsf{Spin}(C)}\mathsf{Trope}_{\kappa}\right)$

where each $\mathsf{Trope}_{\kappa}\cong{\mathbb{P}}^{2}\subset
X={\mathbb{P}}^{3}$ is a plane corresponding to the fixed point
$\mathbb{E}_{\kappa}$. Outside of its $16$ nodes $\mathsf{Kum}$ is simply
tangent to each $\mathsf{Trope}_{\kappa}$ along a smooth conic
$\mathfrak{C}_{\kappa}\subset\mathsf{Trope}_{\kappa}\cong{\mathbb{P}}^{2}$,
and any two trope planes intersect transversally.

###### Proof.

We already described the components of the wobbly locus in the proof of
Theorem 3.6 but for completeness we recall the relevant arguments here. A
wobbly bundle $E$ is either strictly semistable, in which case it corresponds
to a point of $\mathsf{Kum}$, or it is stable but has a nonzero nilpotent
Higgs field. In the second case, the upward limit of the
${\mathbb{C}}^{\times}$ orbit corresponding to that Higgs field is a fixed
point not in $X$, so by Proposition 3.3(a) the limiting Higgs bundle has the
form $\kappa\oplus\kappa^{-1}$ where $\kappa$ is a square-root of
$\omega_{C}$. In terms of the bundle $E$, it means that there is an exact
sequence

$0\rightarrow\kappa^{-1}\rightarrow E\rightarrow\kappa\rightarrow 0.$

Thus $E$ corresponds to an extension class $\xi\in
H^{1}(\kappa^{-2})=H^{1}(\omega_{C}^{-1})\cong H^{0}(\omega_{C}^{\otimes
2})^{\vee}$. The set of bundles corresponding to extension classes like that
is a ${\mathbb{P}}^{2}$. Narasimhan-Ramanan’s discussion [NR69, Proposition
6.1, Theorem 2] shows that the map from the projectivized space of extension
classes into $X={\mathbb{P}}^{3}$ is linear, thus we obtain a linear
$\mathsf{Trope}_{\kappa}\cong{\mathbb{P}}^{2}\subset{\mathbb{P}}^{3}$. These
define the $16$ divisors $\mathsf{Trope}_{\kappa}$. They are different since
the fixed points are different. The intersections
$\mathsf{Trope}_{\kappa}\cap\mathsf{Trope}_{\kappa^{\prime}}$ are therefore
transverse. Note also that, as we will see in Corollary 5.5 below, three
distinct trope planes can not meet along a line, so the intersections between
trope planes are normal crossings in $X$-codimension $2$.

We consider next the condition that an extension of the above form lies in the
Kummer surface. It means that there is a line bundle $U$ of degree $0$ with a
map $U\rightarrow\kappa$ such that the extension splits. We may write
$U=\kappa(-t)$ for a point $t\in C$. Splitting the extension means that the
image of $\xi$ under the map

$H^{1}(\omega_{C}^{-1})\rightarrow H^{1}(\omega_{C}^{-1}(t))\cong
H^{1}({\mathcal{O}}_{C}(-t^{\prime}))$

should vanish, where $t^{\prime}$ is the hyperelliptic conjugate of $t$. This
determines an extension $\xi$ uniquely up to scalars. Furthermore, the
extension determined by $t^{\prime}$ yields an $S$-equivalent bundle since

$\kappa(-t)\otimes\kappa(-t^{\prime})=\omega_{C}(-t-t^{\prime})={\mathcal{O}}_{C}.$

Thus, the map from $C$ to the space of extensions factors through a map from
${\mathbb{P}}^{1}$. Notice that the map from $C$ to $\mathsf{Kum}$ comes by
sending $t$ to the line bundle $\kappa(-t)\in\operatorname{Jac}^{0}(C)$ and
then projecting from the Jacobian to the Kummer.

Let us consider a different plane given by the space of extensions of the form

$0\rightarrow V^{-1}\rightarrow E\rightarrow V\rightarrow 0$

for $V$ some general line bundle of degree $1$. This plane is the space of
bundles such that $H^{0}(V\otimes E)\neq 0$. If that holds for $E$ then it
holds for the polystable bundle in the $S$-equivalence class of $E$, so it
means that we look for the condition $H^{0}(V\otimes\kappa(-t))\neq 0$. We
note that the degree $2$ bundle $V\otimes\kappa$ is general, so it has a
single section with two zeros. If $t$ is one of these zeros, then we get a
solution. The case of a hyperelliptically conjugate point is when the other
piece $H^{0}(V\otimes\kappa(-t^{\prime}))\neq 0$. There are two points in
${\mathbb{P}}^{1}$ that are the images of the zeros of the section of
$V\otimes\kappa$, corresponding to the intersection with this plane. We
conclude that the rational curve image of ${\mathbb{P}}^{1}$ is a conic.

Thus, $\mathsf{Kum}\cap\mathsf{Trope}_{\kappa}$ has reduced scheme equal to a
conic, so it is twice a conic in the plane
$\mathsf{Trope}_{\kappa}\cong{\mathbb{P}}^{2}$. ∎

######  Remark 5.3.

In the classical studies of the geometry of the Kummer surface (see [Kle70,
GH94, NR69, Bea96, Keu97, Dol20, Hud05]), these $16$ planes were known as the
trope planes. Each trope plane passes through $6$ of the $16$ singular points
of the Kummer surface, and each singular point is contained in $6$ of the $16$
trope planes. This configuration was known as the _Kummer $16_{6}$
configuration_. The conics $\mathfrak{C}_{\kappa}$ are called _trope conics_.
We have adopted the ‘trope’ terminology.

######  Proposition 5.4.

Inside one of the trope plane $\mathsf{Trope}_{\kappa}$, the trope conic may
be identified with the hyperelliptic ${\mathbb{P}}^{1}$. The six branch points
in the hyperelliptic ${\mathbb{P}}^{1}$ correspond to the $6$ nodes of
$\mathsf{Kum}$ contained in that trope plane. The $15$ lines of intersection
with the other trope planes, consist of all lines passing through pairs of
branch points (i.e. nodes of $\mathsf{Kum}$ on the trope conic.

###### Proof.

In modular terms, a trope plane corresponds to a choice of a square-root
$\kappa$ of $\omega_{C}$, with the corresponding
$\mathsf{Trope}_{\kappa}\subset X\cong{\mathbb{P}}^{3}$ consisting of all the
bundles that are extensions of the form

$0\rightarrow\kappa^{-1}\rightarrow E\rightarrow\kappa\rightarrow 0.$

Such a bundle $E$ belogs to $\mathsf{Kum}$ if it contains a degree $0$ line
bundle; such would be of the form $L=\kappa(-t)$ with the dual
$L^{-1}=\kappa(-t^{\prime})$ where $t^{\prime}$ is the hyperelliptic conjugate
of $t$. The bundle determines in this way, and is determined by, a point on
the hyperelliptic ${\mathbb{P}}^{1}$. The nodes of the Kummer surface occur
when $L^{-1}\cong L$, thus $t^{\prime}=t$ in other words these correspond to
branch points in the hyperelliptic ${\mathbb{P}}^{1}$. Now, given two
different branch points $\boldsymbol{\mathsf{x}}_{1}$ and
$\boldsymbol{\mathsf{x}}_{2}$ we get
$(\kappa^{\prime})^{-1}:=\kappa(-\boldsymbol{\mathsf{x}}_{1}-\boldsymbol{\mathsf{x}}_{2})$.
This is the dual of another square-root $\kappa^{\prime}$ of $\omega_{C}$. The
set of bundles $E$ that admit non-zero maps from $(\kappa^{\prime})^{-1}$ is
the intersection of $\mathsf{Trope}_{\kappa}$ and
$\mathsf{Trope}_{\kappa^{\prime}}$. This set contains in particular the
bundles containing $\kappa(-\boldsymbol{\mathsf{x}}_{1})$ respectively
$\kappa(-\boldsymbol{\mathsf{x}}_{2})$, i.e. this intersection line passes
through the two branch points $\boldsymbol{\mathsf{x}}_{1}$ and
$\boldsymbol{\mathsf{x}}_{2}$ on the trope conic. We have identified a set of
$15$ lines that is on the one hand the set of intersection lines between
$\mathsf{Trope}_{\kappa}$ and all the other planes, and on the other hand is
the set of lines through pairs of the $6$ branch points. ∎

######  Corollary 5.5.

No three trope planes pass through the same line.

###### Proof.

Recall that we are assuming genericity of the set of $6$ branch points of the
hyperelliptic curve. If there were three trope planes passing through the same
line, then another trope plane would contain a configuration of three lines of
the form
$\mathsf{L}_{\kappa\kappa^{\prime}}=\mathsf{Trope}_{\kappa}\cap\mathsf{Trope}_{\kappa^{\prime}}$
as in the proof of the proposition, that pass through the same point. However,
noting that all smooth plane conics are projectively equivalent to a given
fixed one, our set of $6$ branch points is general on the conic. If three of
the lines $L_{\kappa\kappa^{\prime}}$ were to pass through the same point, we
could jiggle one of the branch points to move one of the three lines away from
that intersection point, so this can not happen under our genericity
hypothesis. ∎

Let ${\mathcal{P}}\subset\mathsf{Higgs}_{0}$ be the fiber of the Hitchin
fibration corresponding to a spectral curve $\pi:{\widetilde{C}}\rightarrow
C$. Notice that, although the Hitchin fibrations are different for degree $0$
and $1$, the base is still the space of quadratic differentials and the
spectral curves are the same. Thus, Proposition 3.1 applies, and
${\widetilde{C}}/C$ is branched over two pairs of hyperelliptically conjugate
points $a,a^{\prime},b,b^{\prime}$.

Then, as we saw in Theorem 3.6 we need to blow up
${\mathcal{P}}={\mathcal{P}}_{2}$ at the $16$ points
$\\{\pi^{*}\kappa\\}_{\kappa\in\mathsf{Spin}(C)}$ to get $Y$ with a map
$f:Y\rightarrow X$. Let $H$ be the hyperplane class on $X$, let
$\boldsymbol{\mathit{F}}$ be its pullback to $Y$, and let
$\boldsymbol{\mathit{E}}=\sum_{\kappa\in\mathsf{Spin}(C)}\boldsymbol{\mathit{E}}_{\kappa}$
be the exceptional divisor (union of $16$ disjoint ${\mathbb{P}}^{2}$’s).

######  Lemma 5.6.

Under the finite morphism $f:Y\to X$ each plane
$\boldsymbol{\mathit{E}}_{\kappa}$ maps isomorphically to the corresponding
plane $\mathsf{Trope}_{\kappa}$ in $X$, the map from $Y$ to $X$ being ramified
with index $2$ there, and this accounts for the ramification over general
points of $\mathsf{Trope}_{\kappa}$.

###### Proof.

We have seen in the proof of Theorem 3.6 and in Proposition 5.2 that the plane
$\mathsf{Trope}_{\kappa}$ is the projectivized set of downward directions at
the fixed point, which is isomorphic to the projectivized normal bundle of the
upward subset emanating from the fixed point. This in turn is isomorphic to
the projectivized normal bundle of the intersection point of the upward set
with ${\mathcal{P}}$, which is the exceptional divisor
$\boldsymbol{\mathit{E}}_{\kappa}$.

Let us recall why the upward subset intersects the Prym in a single point
(this explains why there are $16$ points to blow up, as stated above). Our
fixed point corresponds to a nilpotent Higgs bundle with underlying bundle of
the form $\kappa\oplus\kappa^{-1}$ where $\kappa$ is a square-root of
$\omega_{C}$. The upward subset is the set of Higgs bundles whose underlying
vector bundle is unstable with detabilizing subbundle $L$. To intersect with
the Prym, we look for line bundles $U$ on ${\widetilde{C}}$ of degree $2$
whose norm is $\omega_{C}$, such that there is a map
$\kappa\rightarrow\pi_{*}(U)$ or equivalently $\pi^{*}\kappa\rightarrow U$.
But $\pi^{*}\kappa$ also has degree $2$ on ${\widetilde{C}}$ so
$U=\pi^{*}\kappa$ is the unique solution.

There is an involution of ${\mathcal{P}}$ given by applying the hyperelliptic
involution on ${\widetilde{C}}$ (which covers the hyperelliptic involution on
$C$). Up to translation by $\omega_{C}$, this amounts to doing $-1$ on the
abelian variety, and the $16$ points being blown up are the fixed points.
Normal directions to these points are also fixed, so the
$\boldsymbol{\mathit{E}}_{\kappa}$ are fixed points of the induced involution
of $Y$.

The pullback of a semistable degree $0$ bundle by the hyperelliptic involution
on $C$ is $S$-equivalent to itself. This is known for local systems in the
work of Goldman and Heu-Loray [Gol97, HL19], in particular it applies to
unitary local systems and hence to polystable bundles.

The map $f:Y\rightarrow X$ therefore factors through the quotient by the
involution. But the involution acts by $-1$ on the normal directions, so it
acts by $-1$ on the normal bundle of $\boldsymbol{\mathit{E}}_{\kappa}$,
giving a simple ramification divisor of the map from $Y$ to its quotient by
this involution along $\boldsymbol{\mathit{E}}_{\kappa}$.

We claim that this accounts for the ramification of the map $f:Y\rightarrow X$
along $\boldsymbol{\mathit{E}}_{\kappa}$. Suppose that a point of
$\boldsymbol{\mathit{E}}_{\kappa}$ counts for $m$ points in the fiber of
$f:Y\rightarrow X$ that has degree $8$ (Lemma 4.9). We are claiming that
$m=2$, and have shown in the preceding paragraph that $m\geq 2$. A monodromy
argument (when moving everything around, the planes
$\boldsymbol{\mathit{E}}_{\kappa}$ get interchanged transitively) tells us
that this coefficient $m$ is the same for all
$\boldsymbol{\mathit{E}}_{\kappa}$. Now, $\mathsf{Trope}_{\kappa}$ is a plane
inside $X={\mathbb{P}}^{3}$, so a different component
$\mathsf{Trope}_{\kappa^{\prime}}$ intersects it in a line. Consider yet a
third component $\mathsf{Trope}_{\kappa^{\prime\prime}}$ that intersects this
line in at least one point. This is not a smooth point of the Kummer surface,
since the trope planes are tangent planes to the Kummer surface at points on
the trope conics that are different from the $16$ singular points, and the
tangent plane is unique so in that case the three planes would be the same.
The point might be one of the $16$ singular points of the Kummer surface.
However, that does not always happen. Indeed, the other trope planes intersect
$\mathsf{Trope}_{\kappa}$ in $15$ lines, and the various intersection points
of these lines can not be limited to only the singular points of the Kummer
surface (of which there are $6$ corresponding to the Weierstrass points inside
the ${\mathbb{P}}^{1}$ of the trope conic). So we can choose
$\mathsf{Trope}_{\kappa^{\prime}}$ and
$\mathsf{Trope}_{\kappa^{\prime\prime}}$ to correspond to lines that intersect
somewhere in $\mathsf{Trope}_{\kappa}$ other than a singular point.

Now, the exceptional divisors upstairs in $Y$ are three disjoint planes
$\boldsymbol{\mathit{E}}_{\kappa},\boldsymbol{\mathit{E}}_{\kappa^{\prime}},\boldsymbol{\mathit{E}}_{\kappa^{\prime\prime}}$.
In particular, lying over the intersection point inside $X$, there are at
least three different points in the
$\boldsymbol{\mathit{E}}_{\kappa},\boldsymbol{\mathit{E}}_{\kappa^{\prime}},\boldsymbol{\mathit{E}}_{\kappa^{\prime\prime}}$.
This would give $3m\leq 8$, from which we conclude $m=2$.

We now know that in a fiber over a general point of $\mathsf{Trope}_{\kappa}$,
there is a ramification point in $\boldsymbol{\mathit{E}}_{\kappa}$ counting
for $2$ points in the degree $8$ fiber, plus $6$ other points on $Y$ that are
not in the exceptional divisor, in other words they are in the locus of
$\mathsf{Higgs}_{0}$ whose underlying bundle is stable. An argument similar to
the proof of Lemma 5.13 below shows that for a general point in the Hitchin
base, and over a general point of $\mathsf{Trope}_{\kappa}$, the other $6$
points are unramified. ∎

###  5.2 Computations in degree zero

Recall that by convention we drop terms of degree $\geq 3$ in all expressions.

######  Proposition 5.7.

The Todd classes for the degree $0$ moduli spaces are:

$\displaystyle{\rm td}(TX)$ $\displaystyle=1+2H+11H^{2}/6,$ $\displaystyle
f^{*}{\rm td}(TX)^{-1}$