we see that if, for example, $v_{3}$ is in $U_{0}$, then the first two terms
are zero because $v_{3}\in U_{0}\subseteq\ker\kappa_{y}^{L}$ and the last term
is zero because $v_{3}+{}^{y}v_{3}=2v_{3}\in
U_{0}\subseteq\ker\kappa_{x}^{*}$.

This completes the details showing the extension of $\kappa$ is a Drinfeld
orbifold algebra map for $G$ acting on $U\oplus U_{0}$. Since the restriction
and extension maps are inverses, we obtain the desired bijection. ∎

We now use this result to try to extend $\kappa$ from
$\bigwedge^{2}(\mathfrak{h}^{*}\oplus\mathfrak{h})$ to
$\bigwedge^{2}(W^{*}\oplus W)$. However, the form of the Lie orbifold algebra
maps $\kappa^{L}_{1}+\kappa^{C}_{1}+\kappa^{C}_{\operatorname{ref}}$ for
$S_{n}$ acting on its doubled permutation representation will imply that in
fact $\kappa^{L}_{1}\equiv 0$.

######  Theorem 8.2.

For $S_{n}$ $(n\geq 4)$ acting on
$\mathfrak{h}^{*}\oplus\mathfrak{h}\cong\mathbb{C}^{2n-2}$ by the doubled
standard representation there are no degree-one Lie orbifold algebra maps.

###### Proof.

If there were a Lie orbifold algebra map $\kappa^{L}+\kappa^{C}$ for the
doubled standard representation with
$\kappa^{C}:\bigwedge^{2}(\mathfrak{h}^{*}\oplus\mathfrak{h})\to\mathbb{C}$
and nonzero
$\kappa^{L}:\bigwedge^{2}(\mathfrak{h}^{*}\oplus\mathfrak{h})\to\mathfrak{h}^{*}\oplus\mathfrak{h}$,
then it could be extended as described in Proposition 8.1 to yield a Lie
orbifold algebra map for $S_{n}$ acting on $V=W^{*}\oplus
W\cong\mathbb{C}^{2n}$ via the doubled permutation representation. The
possible forms of such extensions $\kappa$ are controlled by Theorem 5.1. By
imposing first the image constraint
$\operatorname{im}\kappa^{L}\subseteq\mathfrak{h}^{*}\oplus\mathfrak{h}$, then
the extension conditions in Proposition 8.1, and then PBW conditions
(6.1)-(6.4), we find that in fact $\kappa^{L}\equiv 0$.

First, use $x_{i}=\bar{x}_{i}+\tfrac{1}{n}x_{[n]}$ and
$y_{i}=\bar{y}_{i}+\tfrac{1}{n}y_{[n]}$ in (4.2)-(4.5) to write the values of
$\kappa^{L}_{1}$ according to the decomposition
$V\cong\mathfrak{h}^{*}\oplus\iota^{*}\oplus\mathfrak{h}\oplus\iota$:

$\displaystyle\kappa^{L}_{1}(x_{i},x_{j})$
$\displaystyle=a_{1}(\bar{x}_{i}-\bar{x}_{j})+b_{1}(\bar{y}_{i}-\bar{y}_{j}),$
$\displaystyle\kappa^{L}_{1}(y_{i},y_{j})$
$\displaystyle=a_{2}(\bar{x}_{i}-\bar{x}_{j})+b_{2}(\bar{y}_{i}-\bar{y}_{j}),$
$\displaystyle\kappa^{L}_{1}(x_{i},y_{i})$
$\displaystyle=a_{3}\bar{x}_{i}+\tfrac{1}{n}(a_{3}+na_{4}){x}_{[n]}+b_{3}\bar{y}_{i}+\tfrac{1}{n}(b_{3}+nb_{4}){y}_{[n]},$
$\displaystyle\kappa^{L}_{1}(x_{i},y_{j})$
$\displaystyle=a_{5}\bar{x}_{i}+a_{6}\bar{x}_{j}+\tfrac{1}{n}(a_{5}+a_{6}+na_{7}){x}_{[n]}+b_{5}\bar{y}_{i}+b_{6}\bar{y}_{j}+\tfrac{1}{n}(b_{5}+b_{6}+nb_{7}){y}_{[n]}.$

Thus $\operatorname{im}\kappa^{L}\subseteq\mathfrak{h}^{*}\oplus\mathfrak{h}$
implies

$\displaystyle a_{3}+na_{4}$ $\displaystyle=0,$ $\displaystyle\qquad
b_{3}+nb_{4}$ $\displaystyle=0,$ $\displaystyle a_{5}+a_{6}+na_{7}$
$\displaystyle=0,$ $\displaystyle\qquad b_{5}+b_{6}+nb_{7}$ $\displaystyle=0.$

Second, impose the extension conditions
$\kappa^{L}(w_{i},v_{j})=\kappa^{L}(v_{i},v_{j})=0$ for $w_{i}$ in the basis
$\\{x_{i+1}-x_{i},y_{i+1}-y_{i}\mid 1\leq i\leq n-1\\}$ of
$\mathfrak{h}^{*}\oplus\mathfrak{h}$ and $v_{i},v_{j}$ in the basis
$\\{x_{[n]},y_{[n]}\\}$ of
$\iota^{*}\oplus\iota=(\mathfrak{h}^{*}\oplus\mathfrak{h})^{\perp}$. From

$\displaystyle\kappa^{L}(x_{i+1}-x_{i},x_{[n]})$
$\displaystyle=na_{1}(x_{i+1}-x_{i})+nb_{1}(y_{i+1}-y_{i})=0\hbox{ and}$
$\displaystyle\kappa^{L}(y_{i+1}-y_{i},y_{[n]})$
$\displaystyle=na_{2}(x_{i+1}-x_{i})+nb_{2}(y_{i+1}-y_{i})=0$

we obtain $a_{1}=b_{1}=a_{2}=b_{2}=0$. We also require that

$\kappa^{L}(x_{i+1}-x_{i},y_{[n]})=(a_{3}+na_{5}-(a_{5}+a_{6}))(x_{i+1}-x_{i})+(b_{3}+nb_{5}-(b_{5}+b_{6}))(y_{i+1}-y_{i})=0$

and

$\kappa^{L}(x_{[n]},y_{i+1}-y_{i})=(a_{3}+na_{6}-(a_{5}+a_{6}))(x_{i+1}-x_{i})+(b_{3}+nb_{6}-(b_{5}+b_{6}))(y_{i+1}-y_{i})=0,$

and thus all four coefficients are zero. Using the results of the image
constraint to simplify those coefficients yields

$\displaystyle a_{4}-a_{5}-a_{7}$ $\displaystyle=0,$ $\displaystyle\qquad
b_{4}-b_{5}-b_{7}$ $\displaystyle=0,$ $\displaystyle a_{4}-a_{6}-a_{7}$
$\displaystyle=0,$ $\displaystyle\qquad b_{4}-b_{6}-b_{7}$ $\displaystyle=0,$

from which it follows that $a_{5}=a_{6}$ and $b_{5}=b_{6}$. The remaining
extension requirement imposes no further constraints on the parameters because
one verifies $\kappa^{L}(x_{[n]},y_{[n]})=0$ using
$a_{5}+a_{6}+na_{7}=b_{5}+b_{6}+nb_{7}=0$.

To analyze the PBW conditions (6.1)-(6.4) it will help to first observe that
the above constraints $a_{5}=a_{6}$ and $a_{5}+a_{6}+na_{7}=0$ yield that

$\displaystyle a_{5}$ $\displaystyle=a_{6}$
$\displaystyle=-\tfrac{n}{2}a_{7},$ and hence that $\displaystyle a_{4}$
$\displaystyle=a_{5}+a_{7}$ $\displaystyle=-\tfrac{n-2}{2}a_{7}$ and
$\displaystyle a_{3}$ $\displaystyle=-na_{4}$
$\displaystyle=\tfrac{n(n-2)}{2}a_{7},$

with corresponding expressions in terms of $b_{7}$ for $b_{3}$, $b_{4}$,
$b_{5}$, and $b_{6}$. These allow the simplification

$\displaystyle\phi_{1,1}$ $\displaystyle(x_{i},x_{j},y_{k})$
$\displaystyle=[-b_{5}a_{6}+b_{7}(a_{3}-a_{5}-a_{6})](x_{i}-x_{j})+[b_{5}(b_{5}+nb_{7})+b_{7}(b_{3}-b_{5}-b_{6})](y_{i}-y_{j})$
$\displaystyle=\tfrac{n^{2}}{4}a_{7}b_{7}(x_{i}-x_{j})+\tfrac{n^{2}}{4}b_{7}^{2}(y_{i}-y_{j}).$

Similarly,

$\phi_{1,1}(y_{i},y_{j},x_{k})=\tfrac{n^{2}}{4}a_{7}^{2}(x_{i}-x_{j})+\tfrac{n^{2}}{4}a_{7}b_{7}(y_{i}-y_{j}).$

Requiring each of these to be zero forces $a_{7}=b_{7}=0$, and thus
$a_{i}=b_{i}=0$ for $3\leq i\leq 6$ as well. Since we already have
$a_{1}=b_{1}=a_{2}=b_{2}=0$, this proves there are no Lie orbifold algebra
maps for $S_{n}$ acting on the doubled standard subrepresentation
$\mathfrak{h}^{*}\oplus\mathfrak{h}$ with $\kappa^{L}\not\equiv 0$. ∎

For maps with linear part supported only off the identity there is instead a
three-parameter family of Drinfeld orbifold algebra maps that generalize the
commutator relations for the rational Cherednik algebra $H_{0,c}$.

######  Theorem 8.3.

For $S_{n}$ ($n\geq 4$) acting on
$\mathfrak{h}^{*}\oplus\mathfrak{h}\cong\mathbb{C}^{2n-2}$ by the doubled
standard representation, all Drinfeld orbifold algebra maps with nonzero
linear part supported only off the identity have the form
$\kappa^{L}+\kappa^{C}$ defined by

$\displaystyle(\kappa^{L}+\kappa^{C})(\bar{x}_{i},\bar{x}_{j})$
$\displaystyle=\tfrac{n^{2}}{4}(b^{\perp})^{2}\\!\textstyle\sum\limits_{k\neq
i,j}(ijk)-(kji)$
$\displaystyle(\kappa^{L}+\kappa^{C})(\bar{y}_{i},\bar{y}_{j})$
$\displaystyle=\tfrac{n^{2}}{4}(a^{\perp})^{2}\\!\textstyle\sum\limits_{k\neq
i,j}(ijk)-(kji)$
$\displaystyle(\kappa^{L}+\kappa^{C})(\bar{x}_{i},\bar{y}_{i})$
$\displaystyle=-\textstyle\sum\limits_{k\neq
i}\left[\tfrac{n}{2}\left(a^{\perp}(\bar{x}_{i}+\bar{x}_{k})+b^{\perp}(\bar{y}_{i}+\bar{y}_{k})\right)-c\right]\otimes(ik)$
$\displaystyle(\kappa^{L}+\kappa^{C})(\bar{x}_{i},\bar{y}_{j})$
$\displaystyle=\left[\tfrac{n}{2}(a^{\perp}(\bar{x}_{i}+\bar{x}_{j})+b^{\perp}(\bar{y}_{i}+\bar{y}_{j}))-c\right]\otimes(ij)-\tfrac{n^{2}}{4}a^{\perp}b^{\perp}\\!\textstyle\sum\limits_{k\neq
i,j}(ijk)-(kji).$

###### Proof.

Suppose $\kappa=\kappa^{L}+\kappa^{C}$ is a Drinfeld orbifold algebra map for
the doubled standard representation with
$\kappa^{C}:\bigwedge^{2}(\mathfrak{h}^{*}\oplus\mathfrak{h})\to\mathbb{C}S_{n}$
and nonzero
$\kappa^{L}:\bigwedge^{2}(\mathfrak{h}^{*}\oplus\mathfrak{h})\to(\mathfrak{h}^{*}\oplus\mathfrak{h})\otimes\mathbb{C}S_{n}$
supported only off the identity. Extend $\kappa$ as described in Proposition
8.1 to yield a Drinfeld orbifold algebra map for $S_{n}$ acting on
$V=W^{*}\oplus W\cong\mathbb{C}^{2n}$ via the doubled permutation
representation. By Theorem 5.3 the possible forms of such extensions are
$\kappa^{L}_{\operatorname{ref}}+\kappa^{C}_{1}+\kappa^{C}_{\operatorname{ref}}+\kappa^{C}_{\operatorname{3-cyc}}$
satisfying the PBW conditions (7.3) and (7.4).

We start by imposing the condition
$\operatorname{im}\kappa^{L}\subseteq\mathfrak{h}^{*}\oplus\mathfrak{h}$. Use
$\bar{x}_{i}=x_{i}-\tfrac{1}{n}x_{[n]}$,
$\bar{y}_{i}=y_{i}-\tfrac{1}{n}y_{[n]}$,
$\bar{x}_{ij}:=\bar{x}_{i}+\bar{x}_{j}$, and
$\bar{y}_{ij}:=\bar{y}_{i}+\bar{y}_{j}$ to rewrite the nonzero values of
$\kappa^{L}_{(ij)}$ as

$\displaystyle\kappa^{L}_{(ij)}(x_{i},y_{i})$
$\displaystyle=\kappa^{L}_{(ij)}(x_{j},y_{j})=-\kappa^{L}_{(ij)}(x_{i},y_{j})=-\kappa^{L}_{(ij)}(x_{j},y_{i})$
$\displaystyle=ax_{ij}+a^{\perp}x_{ij}^{\perp}+by_{ij}+b^{\perp}y_{ij}^{\perp}$
$\displaystyle=(a-a^{\perp})x_{ij}+a^{\perp}x_{[n]}+(b-b^{\perp})y_{ij}+b^{\perp}y_{[n]}$
$\displaystyle=(a-a^{\perp})\bar{x}_{ij}+\tfrac{1}{n}(2a+(n-2)a^{\perp})x_{[n]}+(b-b^{\perp})\bar{y}_{ij}+\tfrac{1}{n}(2b+(n-2)b^{\perp})y_{[n]}.$

This shows $2a+(n-2)a^{\perp}=2b+(n-2)b^{\perp}=0$ or

(8.1) $a-a^{\perp}=-\frac{n}{2}a^{\perp}\quad\hbox{ and }\quad
b-b^{\perp}=-\frac{n}{2}b^{\perp}$

and substituting these into (7.3) and (7.4) yields that

$a(\alpha-\beta)=a^{\perp}(\alpha-\beta)=b(\alpha-\beta)=b^{\perp}(\alpha-\beta)=0.$

Thus since $\kappa^{L}\not\equiv 0$ we must also have $\alpha=\beta$ in
$\kappa^{C}_{1}$.

Next consider conditions arising from the definition of the extensions in
Proposition 8.1 by requiring $\kappa^{C}(w,v)=\kappa^{C}(u,v)=0$ for all $w$
in the basis $\\{x_{i+1}-x_{i},y_{i+1}-y_{i}\mid 1\leq i\leq n-1\\}$ of
$\mathfrak{h}^{*}\oplus\mathfrak{h}$ and $u,v$ in the basis
$\\{x_{[n]},y_{[n]}\\}$ of
$\iota^{*}\oplus\iota=(\mathfrak{h}^{*}\oplus\mathfrak{h})^{\perp}$.

For $\kappa^{C}_{1}$, since
$\kappa^{C}_{1}(x_{i},x_{j})=\kappa^{C}_{1}(y_{i},y_{j})=0$ and
$\kappa^{C}_{1}(x_{i},y_{[n]})=\kappa^{C}_{1}(x_{[n]},y_{i})=\alpha+(n-1)\beta$
for any $1\leq i,j\leq n$, the only extension condition that is not
automatically satisfied is
$\kappa^{C}_{1}(x_{[n]},y_{[n]})=n(\alpha+(n-1)\beta)=0$. Together with
$\alpha=\beta$ this forces $\alpha=\beta=0$ and thus $\kappa^{C}_{1}\equiv 0$.

For $\kappa^{*}_{\operatorname{ref}}$ and $\kappa^{C}_{\operatorname{3-cyc}}$,
the definitions of $\kappa^{L}_{g}$ and $\kappa^{C}_{g}$ when $g$ is a
transposition and of $\kappa^{C}_{g}$ when $g$ is a $3$-cycle yield that

(8.2) $\kappa^{*}_{g}\left(v,\textstyle\sum\limits_{h\in\langle
g\rangle}{}^{h}v\right)=\kappa^{*}_{g}\left(v,\textstyle\sum\limits_{h\in\langle
g\rangle}{}^{h}v^{*}\right)=0$

for all $v\in\\{x_{1},\dots,x_{n},y_{1},\dots,y_{n}\\}$. This in turn implies
that all of the extension conditions hold for
$\kappa^{*}_{\operatorname{ref}}$ and $\kappa^{C}_{\operatorname{3-cyc}}$,
yielding no further constraints on parameters.

We now evaluate $\kappa^{L}_{\operatorname{ref}}$,
$\kappa^{C}_{\operatorname{ref}}$, and $\kappa^{C}_{\operatorname{3-cyc}}$ at
pairs of vectors from
$\\{\bar{x}_{1},\dots,\bar{x}_{n},\bar{y}_{1},\dots,\bar{y}_{n}\\}$. For $g$ a
transposition, $i$ an index moved by $g$, $\bar{v}$ a vector in
$\\{\bar{x}_{i},{}^{g}\bar{x}_{i},\bar{y}_{i},{}^{g}\bar{y}_{i}\\}$, and $*=L$
or $*=C$, we have $\kappa^{*}_{g}(\bar{v},{}^{g}\bar{v})=0$ and we use (8.2)
to observe that for $1\leq i\leq n$,

$\displaystyle\kappa^{*}_{g}(\bar{x}_{i},\bar{y}_{i})=-\kappa^{*}_{g}(\bar{x}_{i},{}^{g}\bar{y}_{i})$
$\displaystyle=\kappa^{*}_{g}(x_{i},y_{i})-\tfrac{1}{n}\kappa^{*}_{g}(x_{i},y_{[n]})-\tfrac{1}{n}\kappa^{*}_{g}(x_{[n]},y_{j})+\tfrac{1}{n^{2}}\kappa^{*}_{g}(x_{[n]},y_{[n]})$
$\displaystyle=\kappa^{*}_{g}(x_{i},y_{i}).$

It then follows from (8.1) that

$\displaystyle\kappa^{L}_{g}(\bar{x}_{i},\bar{y}_{i})$
$\displaystyle=-\kappa^{L}_{g}(\bar{x}_{i},{}^{g}\bar{y}_{i})=-\tfrac{n}{2}a^{\perp}\bar{x}_{ij}-\tfrac{n}{2}b^{\perp}\bar{y}_{ij},\hbox{
and }$ $\displaystyle\kappa^{C}_{g}(\bar{x}_{i},\bar{y}_{i})$
$\displaystyle=-\kappa^{C}_{g}(\bar{x}_{i},{}^{g}\bar{y}_{i})=c.$

For $g$ a three-cycle, by the orbit property in (8.2) and by (8.1) we see that

$\displaystyle\kappa^{C}_{g}(\bar{v},\bar{v}^{*})$ $\displaystyle=0$
$\displaystyle\kappa^{C}_{g}(\bar{v},{}^{g}\bar{v})$
$\displaystyle=\begin{cases}\frac{n^{2}}{4}(a^{\perp})^{2}&\text{if $v\in
W$},\\\ \frac{n^{2}}{4}(b^{\perp})^{2}&\text{if $v\in
W^{*}$},\end{cases}\hbox{ and}$
$\displaystyle\kappa^{C}_{g}({}^{g}\\!\bar{v},\bar{v}^{*})=-\kappa^{C}_{g}(\bar{v},{}^{g}\\!\bar{v}^{*})$
$\displaystyle=\tfrac{n^{2}}{4}a^{\perp}b^{\perp}.$

These components produce the given definition of $\kappa^{L}+\kappa^{C}$. ∎

######  Remark 8.4.

In the case $\kappa^{L}_{1}=\kappa^{L}_{\operatorname{ref}}\equiv 0$ then also
$\kappa^{C}_{\operatorname{3-cyc}}\equiv 0$, $\alpha=-(n-1)\beta$, and the
restriction of the constant $2$-cochain
$\kappa^{C}=\kappa^{C}_{1}+\kappa^{C}_{\operatorname{ref}}$ to
$\bigwedge^{2}(\mathfrak{h}^{*}\oplus\mathfrak{h})$ is given by

$\displaystyle\kappa^{C}_{1}(\bar{x}_{i},\bar{x}_{j})$
$\displaystyle=\kappa^{C}_{1}(\bar{y}_{i},\bar{y}_{j})=0,$
$\displaystyle\qquad\kappa^{C}_{1}(\bar{x}_{i},\bar{y}_{i})$
$\displaystyle=-(n-1)\beta,$
$\displaystyle\qquad\kappa^{C}_{1}(\bar{x}_{i},\bar{y}_{j})$
$\displaystyle=\beta,$ $\displaystyle\kappa^{C}_{g}(\bar{x}_{i},\bar{x}_{j})$
$\displaystyle=\kappa^{C}_{g}(\bar{y}_{i},\bar{y}_{j})=0,$
$\displaystyle\qquad\kappa^{C}_{g}(\bar{x}_{i},\bar{y}_{i})$
$\displaystyle=c,$
$\displaystyle\qquad\kappa^{C}_{g}(\bar{x}_{i},\bar{y}_{j})$
$\displaystyle=-c,$

where $\beta,c\in\mathbb{C}$, $g$ is a transposition, and $1\leq i\neq j\leq
n$. This corresponds to the rational Cherednik algebra $H_{n\beta,c}$.

In the theory of rational Cherednik algebras, $H_{t,c}$, for the symmetric
group, a natural isomorphism between $H_{t,c}$ and $H_{\lambda t,\lambda c}$
when $\lambda\in\mathbb{C}^{\times}$ means that only two distinct cases need
be considered, $t\neq 0$ and $t=0$. Theorems 8.2 and 8.3 show that in the
first case there are no further deformations in polynomial degree one with the
linear part of the parameter map supported only on the identity while there is
a three-parameter family of such deformations in the second case with the
linear part of the parameter map supported only off the identity.

######  Remark 8.5.

What if $\kappa^{L}$ were supported both on and off the identity? The
specializations of parameter values for $\kappa^{L}_{1}$ in part (1) and for
$\kappa^{L}_{\operatorname{ref}}$ in part (3) of Theorem 5.4 on the combined
lift of $\kappa^{L}_{1}+\kappa^{L}_{\operatorname{ref}}$ means that

$\displaystyle\kappa^{L}_{1}(x_{i},y_{i})$
$\displaystyle=\kappa^{L}_{1}(x_{i},y_{j})=a_{4}x_{[n]}+b_{4}y_{[n]},\quad\hbox{
and}$ $\displaystyle\kappa^{L}_{(ij)}(x_{i},y_{i})$
$\displaystyle=-\kappa^{L}_{(ij)}(x_{i},y_{j})=-\tfrac{n}{2}a^{\perp}(\bar{x}_{i}+\bar{x}_{j})-\tfrac{n}{2}b^{\perp}(\bar{y}_{i}+\bar{y}_{j}).$

But then
$\operatorname{im}(\kappa^{L}_{1}+\kappa^{L}_{\operatorname{ref}})\subseteq\mathfrak{h}^{*}\oplus\mathfrak{h}$
would require $a_{4}=b_{4}=0$ so $\kappa^{L}_{1}\equiv 0$. This combined with
part (2) of Theorem 5.4 shows there is no Drinfeld orbifold algebra map for
$S_{n}$ on $\mathfrak{h}^{*}\oplus\mathfrak{h}$ with linear part supported
both on and off the identity which extends to a map $\kappa$ of the form in
Theorem 5.4. But since Theorem 5.4 is not exhaustive, it is not clear whether
there exist such maps in general and this is one possible area for further
work.

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