characteristic polynomial when $n$ is small, by first using Theorem 1.1. For
example for $n=3$, $a=\alpha{I}+\bar{a}(\lambda)$, $\alpha\neq 0$, it gives

$\begin{array}[]{|c|c|c|c|}\hline\cr\lambda&[1^{3}]&[1,2]&[3]\\\ \hline\cr
K_{\lambda}&q^{3}K^{3}+(q^{5}+2q^{4})(q-1)K&q^{3}K^{3}+q^{4}(q-2)K&q^{3}K^{3}-2q^{4}K\\\
\hline\cr\end{array}.$

The first non-trivial example when $\lambda^{\prime\prime\prime}$ appears is
$[2^{2}]\vdash 4$, for which

$K_{[2^{2}]}=q^{3}KK_{[1,2]}-q^{7}K_{[2]}-q^{6}(q-1)(K_{[2]}-K_{[1^{2}]})=q^{6}K^{4}+q^{7}(q-3)K^{2}+q^{8}(q^{2}-q+1).$

To see a more intricate situation we illustrate the algorithm for $n=6$ and
$\lambda=[1,2,3]$, when we have

$K_{[1,2,3]}=q^{5}KK_{[1,2^{2}]}-q^{10}K_{[1^{2},2]}.$

Furthermore

$K_{[1,2^{2}]}=q^{4}K_{[1^{2},2]}-2q^{8}K_{[1,2]}$

and so on.

There are many families where the recursion may be stated in simple terms, for
example if there is only one block, $\lambda=[n]$, when we have

$K_{[n]}=Kq^{n-1}K_{[n-1]}-q^{2n-2}K_{[n-2]}.$

From this one can get a closed formula for $K_{[n]}$ see subsection 5.5 below.

Theorem 5.1 is an easy corollary of the following two propositions. As usual
we assume $\alpha\neq 0$, and $a=\alpha{I}+\bar{a}(\lambda)$ whith
$\varepsilon_{j}\in\\{0,1\\}$ as in (21) with $\varepsilon_{n-1}=1$ and use
$a^{\prime\prime}$ to denote $a^{\prime\prime}_{\not{k},\not{n}}$.

######  Proposition 5.2.

Let
$Z=\left\\{z=\sum_{j=i+1}^{l-1}\xi_{j}\mathbf{e}_{k-1,N_{j}-1}\,|\,\xi_{j}\in\mathbb{F}_{q}\right\\}$.
In the above notation

  1. 1.

$a^{\prime\prime}+z$ and $a^{\prime\prime}$ are conjugate over
$\mathbb{F}_{q}$.

  2. 2.

$a^{\prime\prime}+z+\mathbf{e}_{k-1,n-2}$ and
$a^{\prime\prime}+\mathbf{e}_{k-1,n-2}$ are conjugate over $\mathbb{F}_{q}$.

As an immediate corollary of (27) we get that $\sum_{x\in
X_{k}}\psi(ax+x^{-1})=0$, unless $k=N_{i}$ for some $k<l$ such that
$n_{k}=n_{l}$ and then

$\sum_{x\in
X_{k}}\psi(ax+x^{-1})=q^{2n+k-l-3}(q-1)\Big{(}K_{n-2}(a^{\prime\prime})-K_{n-2}(a^{\prime\prime}+\mathbf{e}_{k-1,n-2})\Big{)}$

The second step in the proof is the following

######  Proposition 5.3.

  1. 1.

If $n_{i}<n_{l}$ then $a^{\prime\prime}+\mathbf{e}_{k-1,n-2}$ and
$a^{\prime\prime}$ are conjugate over $\mathbb{F}_{q}$.

  2. 2.

If $n_{i}=n_{l}$, $a^{\prime\prime}+\mathbf{e}_{k-1,n-2}$ is conjugate to
$a^{\prime\prime\prime}$, where $a^{\prime\prime\prime}$ is built from the
partition $\lambda^{\prime\prime\prime}$ as in (21).

While not needed, we remark that the proposition remains true over
$\mathbb{Z}$.

The proof of the claims in the propositions will use the linear transformation
interpretation form subsection 2.6. We start with an easy observation.

######  Lemma 5.4.

Let $V_{A}=\langle v_{0}\rangle\oplus...\oplus\langle
v_{l}\rangle\simeq\mathcal{C}_{n_{1}}\oplus...\oplus\mathcal{C}_{n_{l}}$. If
$v\in V_{A}$ is such that $A^{k}v=0$ for $k<n_{0}$, then $V_{A}=\langle
v_{0}+v\rangle\oplus...\oplus\langle v_{l}\rangle$ as well.

###### Proof.

One easily checks that $\langle v_{1}+v\rangle\cap\big{(}\langle
v_{2}\rangle\oplus...\oplus\langle v_{l}\rangle\big{)}=\\{0\\}$ and that
$\langle v_{1}+v\rangle+\big{(}\langle v_{2}\rangle\oplus...\oplus\langle
v_{l}\rangle\big{)}=V_{A}$. ∎

######  Remark 8.

It follows that there is an isomorphism $\phi:V_{A}\to V_{A}$ which is trivial
on $\langle v_{2}\rangle\oplus...\oplus\langle v_{l}\rangle$ and extends
$v_{1}\mapsto v_{1}+v$. Clearly, it satisfies $A\phi=\phi A$.

The question for us is to determine how a module structure given by $A:V\to V$
changes if $A$ is perturbed by another map $Z:V\to V$. For example Proposition
5.2 and (i) of Proposition 5.3 are an easy consequence of the following lemma.

######  Lemma 5.5.

Assume that $V_{A}\simeq\langle v_{0}\rangle\oplus\langle
v_{1}\rangle...\oplus\langle
v_{l}\rangle\simeq\mathcal{C}_{n_{0}}\oplus\mathcal{C}_{n_{1}}\oplus...\oplus\mathcal{C}_{n_{l}}$
and that $Z:V\to V$ is such that for $i=1,...,l$

$Z(A^{j}v_{i})=0\quad\text{ for }j=0,...,n_{i}-1$

and that moreover for $i=0$ we have

$\displaystyle Z(A^{j}v_{0})$ $\displaystyle=0\quad\text{ for
}j=0,...,n_{0}-2,$ $\displaystyle Z(A^{n_{0}-1}v_{0})$
$\displaystyle=\sum_{i=1}^{l}\xi_{i}A^{n_{i}-1}v_{i}.$

If $n_{0}<n_{i}$ for all $i=1,...,l$ then $V_{A+Z}\simeq V_{A}$ as
$\mathbb{F}_{q}[T]$-modules.

###### Proof.

Let $v=\sum_{i=1}^{l}\xi_{i}A^{n_{i}-n_{0}}v_{i}$. Clearly $A^{n_{0}}v=0$, and
so by the Lemma 5.4 there is an $A$-linear isomorphism $\phi$, for which
$\phi(v_{0})=v_{0}-v$. One easily checks that $Z\circ\phi=0$ and so $\phi$
provides the claim isomorphism. ∎

######  Lemma 5.6.

Assume that $V_{A}\simeq\langle v_{0}\rangle\oplus\langle
v_{1}\rangle\simeq\mathcal{C}_{m}\oplus\mathcal{C}_{m}$ and that $Z:V\to V$ is
such that

$\displaystyle Z(A^{j}v_{1})$ $\displaystyle=0\quad\text{ for }j=0,...,m-1,$
$\displaystyle Z(A^{j}v_{0})$ $\displaystyle=0\quad\text{ for }j=0,...,m-2,$
$\displaystyle Z(A^{m-1})v_{0}$ $\displaystyle=\xi A^{m-1}v_{1}$

for some $\xi\in\mathbb{F}_{q}^{*}$. Then
$V_{A+Z}\simeq\mathcal{C}_{m-1}\oplus\mathcal{C}_{m+1}$ as
$\mathbb{F}_{q}[T]$-modules.

###### Proof.

It is easy to see that $(A+Z)^{j}v_{0}=A^{j}v_{0}$, for $j=0,...,m-1$ and that
$(A+Z)^{m}v_{0}=\xi A^{m-1}v_{1}$. Moreover if we replace $v_{1}$ by
$v_{1}^{\prime}=v_{1}-\frac{1}{\xi}v_{0}$, then $(A+Z)^{m-1}v_{1}^{\prime}=0$.
∎

###### The Proofs of Proposition 5.2 and 5.3.

The two lemmas above give exactly this. ∎

###  5.3 Examples for the Kloosterman sums over Borel Bruhat cells

Let $B$ be the standard Borel subgroup of invertible upper triangular
matrices, $w\in W$ an element of the Weyl group, and $C_{w}=BwB$. We will
consider here the sums

$K_{n}^{(w)}(a)=\sum_{x\in C_{w}}\psi(ax+x^{-1}),$

where $a=\alpha{I}+\bar{a}$, with $\bar{a}\in\bar{U}$, where $\bar{U}$ is the
set of strictly upper triangular matrices. We will first comment on the nature
on these sums and then derive some of the properties that will be used below
in the section on purity.

As in the proof of Theorem 1.3 in subsection 2.4 one can show that these sums
satisfy a recursion that connects them to similar sums of rank $n-1$ or $n-2$,
depending on whether $w(n)=n$ or $w(n)<n$. To see this, recall from
Proposition 3.14 that

$K_{n}^{(w)}(a)=q^{n_{a}+n_{o}}K_{1}(\alpha)^{f}S_{n}^{(w)}(\overline{a}),$
(51)

where $n_{o}$ is the number of involution pairs in $w$, $f$ is the number of
fixed points of $w$ and where the auxiliary sum is given by

$S_{n}^{(w)}(\overline{a})=\sum_{v,{d,u_{1}}}\psi(\bar{a}^{v}wdu).$

with $v\in U^{\flat},u\in U^{\sharp}_{b}$, $d\in
D(w)=\\{\sum_{i<w(i)}t_{i}\mathbf{e}_{i,i}\,|\,t_{i}\in\mathbb{F}_{q}^{*}\\}$
and $\overline{a}^{v}=v^{-1}\overline{a}v$. The recursion then proceeds on the
sum $S_{n}^{(w)}(a)$. For example when $w(n)=k<n$ we again let
$m^{\prime\prime}=m^{\prime\prime}_{\not{k},\not{n}}$ denote the matrix one
gets by deleting the $k$-th and $n$-th rows and columns of an $n\times n$
matrix $m$. To describe the set of perturbations $Z\subset M_{n-2}$ that arise
in the reduction, we let $\overline{a}_{(i)}$ and $\overline{a}^{(i)}$ denote
the $i$-th row, respectively the $i$th column, of $\overline{a}$ and consider
the set

$Y=\\{y\in\mathbb{F}_{q}^{n}\,|\,y_{i}=0\text{ for }i\leq k,\text{ and
}y\overline{a}=\overline{a}_{(k)}\\}.$

Using this notation we have the following proposition whose proof goes along
the lines of Proposition 2.8 and is omitted here.

######  Proposition 5.7.

$S_{n}^{(w)}(\overline{a})=q^{n-k-1}\sum_{y\in
Y}S_{n-2}^{(w^{\prime\prime})}(\overline{a}^{\prime\prime}+(\overline{a}^{(k)}y)^{\prime\prime})\sum_{t\in\mathbb{F}_{q}^{*}}\varphi(ty\overline{a}^{(n)}).$
(52)

Note that the set $Y$ may be empty, and then so is the set of perturbations
which arise from the collection of rank 1 matrices

$Z=\\{\overline{a}^{(k)}y\in M_{n}\,|\,y\in Y\\},$

in which case the sum is interpreted as 0.

What makes the sums $S_{n}^{(w)}(\overline{a})$ harder to deal with is that as
functions of $a$ they are no longer invariant under conjugation by
$\mathrm{GL}_{n}$. They are invariant under conjugation by elements of $B$ in
virtue of (51) and the fact that

$K_{n}^{(w)}(a)=K_{n}^{(w)}(b^{-1}ab)$

for any $b\in B$, since $b^{-1}C_{w}b=C_{w}$. However the $B$-orbits in the
set $\bar{U}$ of strictly upper triangular matrices under the adjoint action
are not well-understood. It is easy to see that one can no longer straighten
out partitions into a non-decreasing order, or even assume that an orbit is
represented by a matrix in Jordan normal form. For example
$\\{t\mathbf{e}_{1,3}\,|\,t\neq 0\\}$ is one of the orbits in the $3\times 3$
case. When $n=6$, there is even a one-parameter family of orbits, found by
Kashin [Ka90]. In general a full description of the orbits is hard even in low
ranks [BuHe87, HiRo97].

Returning to the sums $K_{n}^{(w)}(a)$ it is still quite likely that these can
be expressed as polynomials in $q$ and $K$, independently of the
characteristic $p$ since the perturbations arising in the reduction
calculations are of a very special nature. In what follows we will provide
some low rank examples, when the independence is easy to establish directly.
We will do this by stating certain special cases when the above reduction is
sufficient, most importantly the case when $w=(i\,j)$. There are a number of
other special cases when the reduction for $S_{n}^{(w)}(a)$ can be treated in
a simple manner, for example when $w(n)=1$ or $n-1$.

Again we assume that $a=\alpha I+\bar{a}$ is of the form as in (8),
$\bar{a}=\sum_{j=1}^{n-1}\varepsilon_{j}\mathbf{e}_{j,j+1}$, and
$\lambda\vdash n$ is the partition corresponding to $a$. Denote

$K_{\lambda}^{(w)}(\alpha)=\sum_{x\in C_{w}}\psi(ax+x^{-1}).$

######  Theorem 5.8.

Let $i<j$ and $(ij)\in W$ and $\alpha\neq 0$. Then

$K^{(ij)}_{\lambda}(\alpha)=q^{n(n-1)/2}K^{n-2}\begin{cases}(q-1)q,&\mathrm{if~{}}j=i+1\mathrm{~{}and~{}}\varepsilon_{i}=0\\\
-q,&\mathrm{if~{}}j=i+1\mathrm{~{}and~{}}\varepsilon_{i}=1\\\
(q-1)q^{j-i-d},&\mathrm{if~{}}j>i+1\mathrm{~{}and~{}}\varepsilon_{i}=\varepsilon_{j-1}=0\end{cases}$

where $d=|\\{k\in\mathbb{N}|i<k<j-1\mathrm{~{}and~{}}\varepsilon_{k}\neq
0\\}|$.

If $j>i+1$ and either $\varepsilon_{i}$ or $\varepsilon_{j-1}\neq 0$ then
$K^{(ij)}_{\lambda}(\alpha)=0$.

This allows us to compute the Bruhat cell polynomials for $n\leq 3$:

######  Corollary 5.9.

$n=2$

$\begin{array}[]{|r|c|c|}\hline\cr\lambda\setminus w&(12)&{I}\\\
\hline\cr[1,1]&q^{2}(q-1)&qK^{2}\\\ \hline\cr[2]&-q^{2}&qK^{2}\\\
\hline\cr\end{array}$

$n=3$

$\begin{array}[t]{|r|c|c|c|c|}\hline\cr\lambda\setminus
w&(13)&(12)&(23)&{I}\\\
\hline\cr[1,1,1]&q^{5}(q-1)K&q^{4}(q-1)K&q^{4}(q-1)K&q^{3}K^{3}\\\
\hline\cr[2,1]&0&q^{4}(q-1)K&-q^{4}K&q^{3}K^{3}\\\
\hline\cr[1,2]&0&-q^{4}K&q^{4}(q-1)K&q^{3}K^{3}\\\
\hline\cr[3]&0&-q^{4}K&-q^{4}K&q^{3}K^{3}\\\ \hline\cr\end{array}$

With a little more work one can calculate all the Bruhat cell polynomials for
$n=4$. We summarize the result in the following table:

######  Corollary 5.10.

$\begin{array}[]{|r|c|c|c|c|c|}\hline\cr\lambda\setminus
w&(14)(23)&(13)(24)&(12)(34)&(14)&(13)\\\
\hline\cr[1,1,1,1]&q^{10}(q-1)^{2}&q^{9}(q-1)^{2}&q^{8}(q-1)^{2}&q^{9}(q-1)K^{2}&q^{8}(q-1)K^{2}\\\
\hline\cr[2,1,1]&0&0&-q^{8}(q-1)&0&0\\\
\hline\cr[1,2,1]&-q^{9}(q-1)&0&q^{8}(q-1)^{2}&q^{8}(q-1)K^{2}&0\\\
\hline\cr[1,1,2]&0&0&-q^{8}(q-1)&0&q^{8}(q-1)K^{2}\\\
\hline\cr[3,1]&0&0&-q^{8}(q-1)&0&0\\\ \hline\cr[2,2]&0&q^{9}(q-1)&q^{8}&0&0\\\
\hline\cr[1,3]&0&0&-q^{8}(q-1)&0&0\\\ \hline\cr[4]&0&0&q^{8}&0&0\\\
\hline\cr\end{array}$ $\begin{array}[]{|r|c|c|c|c|c|}\hline\cr\lambda\setminus
w&(24)&(12)&(23)&(34)&{I}\\\
\hline\cr[1,1,1,1]&q^{8}(q-1)K^{2}&q^{7}(q-1)K^{2}&q^{7}(q-1)K^{2}&q^{7}(q-1)K^{2}&q^{6}K^{4}\\\
\hline\cr[2,1,1]&q^{8}(q-1)K^{2}&-q^{7}K^{2}&q^{7}(q-1)K^{2}&q^{7}(q-1)K^{2}&q^{6}K^{4}\\\
\hline\cr[1,2,1]&0&q^{7}(q-1)K^{2}&-q^{7}K^{2}&q^{7}(q-1)K^{2}&q^{6}K^{4}\\\
\hline\cr[1,1,2]&0&q^{7}(q-1)K^{2}&q^{7}(q-1)K^{2}&-q^{7}K^{2}&q^{6}K^{4}\\\
\hline\cr[3,1]&0&-q^{7}K^{2}&-q^{7}K^{2}&q^{7}(q-1)K^{2}&q^{6}K^{4}\\\
\hline\cr[2,2]&0&-q^{7}K^{2}&q^{7}(q-1)K^{2}&-q^{7}K^{2}&q^{6}K^{4}\\\
\hline\cr[1,3]&0&q^{7}(q-1)K^{2}&-q^{7}K^{2}&-q^{7}K^{2}&q^{6}K^{4}\\\
\hline\cr[4]&0&-q^{7}K^{2}&-q^{7}K^{2}&-q^{7}K^{2}&q^{6}K^{4}\\\
\hline\cr\end{array}$

Note that the polynomials are not merely a permutation for different
rearrangements of a partition, see for example the case $\lambda=[1,2,1]$ and
$w=(14)(23)$.

We finish this section by giving the cell polynomials for the full block case
for general $n$. Let $w=(i_{1},j_{1})(i_{2},j_{2})\dots(i_{r},j_{r})\in W$
such that $i_{k}<j_{k}$ for any $k$. Then

$K^{w}_{[n]}(\alpha)=\begin{cases}(-1)^{r}K^{n-2r}q^{n(n-1)/2+r},&\mathrm{if~{}}j_{k}-i_{k}=1\mathrm{~{}for~{}any~{}}k\\\
0,&\mathrm{otherwise}\end{cases}$

Since this is merely for illustration we only give a sketch of the argument.
For those $w\in W$ such that $K^{w}_{n}(\alpha)=0$ one can find $(i,j)\in I$
such that $\operatorname{tr}(v^{-1}avwdu)$ is nonconstant and linear in
$v_{i,j}$.

###  5.4 Regular semisimple matrices

We have seen that for an $n\times n$ matrix $a$, whose characteristic
polynomials $P_{a}$ have no multiple roots the cohomology associated to the
Kloosterman sum $K_{n}(a)$ is pure. Assume now that this characteristic
polynomial $P_{a}$ is irreducible over $\mathbb{F}_{q}$. Let
$\alpha\in\mathbb{F}_{q^{n}}$ be an eigenvalue of $a$, $P_{a}(\alpha)=0$. The
argument in subsection 3.5 shows that over $\mathbb{F}_{q^{n}}$

$H^{\bullet}=H^{\bullet}_{c}(\mathrm{GL}_{n},x\mapsto\operatorname{tr}(ax+x^{-1}))=\left(\bigotimes_{i=1}^{n(n-1)/2}H^{\bullet}_{c}(\mathbb{A}^{1},0)\right)\otimes\left(\bigotimes_{j=1}^{n}H^{\bullet}_{c}(\mathbb{A}^{1}\setminus\mathbb{A}^{0},f_{\alpha})\right)$

where for $x\in(\mathbb{A}^{1}\setminus\mathbb{A}^{0})(\mathbb{F}_{q^{n}})$,
$f_{\alpha}(x)=\alpha x+x^{-1}$ that corresponds to the scalar Kloosterman sum
$K_{1}(\alpha,\mathbb{F}_{q^{n}})=\lambda_{1}+\lambda_{2}$. Here $\lambda_{1}$
and $\lambda_{2}=\overline{\lambda}_{1}$ are the
$\mathbb{F}_{q^{n}}$-Frobenius eigenvalues on
$H^{\bullet}_{c}(\mathbb{A}^{1}\setminus\mathbb{A}^{0},f_{\alpha})$.

It is then clear that over $\mathbb{F}_{q^{n}}$ the Frobenius eigenvalues on
the $2^{n}$-dimensional space
$\bigotimes_{j=1}^{n}H^{1}_{c}(\mathbb{A}^{1}\setminus\mathbb{A}^{0},f_{\alpha})$
are of the form $(\prod_{i\in I}\lambda_{1})(\prod_{i\notin
I}\lambda_{2})=\lambda_{1}^{|I|}\lambda_{2}^{n-|I|}$, where
$I\subset\\{1,...,n\\}$. (Therefore each of
$\lambda_{1}^{|I|}\lambda_{2}^{n-|I|}$ has multiplicity $\binom{n}{j}$.) If we
fix some $n$-th roots of the $\lambda_{i}$, say $\eta_{i}^{n}=\lambda_{i}$
then we have that the Frobenius eigenvalues on $H^{n^{2}}$ are of the form
$\zeta_{I}q^{n(n-1)/2}\eta_{1}^{|I|}\eta_{2}^{n-|I|}$ where again
$I\subset\\{1,...,n\\}$, and the $\zeta_{I}$ are $n$-th roots of unity,
$\zeta_{I}^{n}=1$ for all $I$. It is natural to make the following

######  Conjecture 5.11.

If $p$ is large enough, and the characteristic polynomial $P_{a}$ is
irreducible over $\mathbb{F}_{q}$ then

$K_{n}(a,\mathbb{F}_{q})=(-1)^{n+1}q^{n(n-1)/2}K_{1}(\alpha,\mathbb{F}_{q^{n}}).$

The conjecture would follow if for $I=\emptyset$ and $\\{1,...,n\\}$ the
$\mathbb{F}_{q}$-Frobenius eigenvalues were
$(-1)^{n+1}q^{n(n-1)/2}\lambda_{1},(-1)^{n+1}q^{n(n-1)/2}\lambda_{2}$ and the
others canceled after summing. For example when $n=3$, this can happen if the
eigenvalues $\mu_{i}$ for $1\leq i\leq 8$ are the eight summands in the
expansion of the product
$(\eta_{1}+\eta_{2})(\omega\eta_{1}+\omega^{2}\eta_{2})(\omega^{2}\eta_{1}+\omega\eta_{2})q^{3}$
for $\omega^{3}=1$. when it leads to

$K_{n}(a,\mathbb{F}_{q})=\sum_{i=1}^{8}\mu_{i}=q^{3}(\lambda_{1}+\lambda_{2})=(-1)^{4}q^{3}K_{1}(\alpha,\mathbb{F}_{q^{3}}),$

exactly as desired.

The conjecture is partly based on the observation that if we let
$K=\mathbb{F}_{q}[a]\subset M_{n}$, then $K$ is a field naturally isomorphic
to $\mathbb{F}_{q^{n}}$. $K$ acts on $M_{n}$ by left multiplication and it is
easy to describe the $K$-algebra that arises for any $n$. We will use this
below to handle the case $n=2$, but such elementary methods get cumbersome and
are unlikely to give a proof, or even offer any hindsight already for $n=3$.

We now give a few numerical examples for $M_{n}(\mathbb{F}_{p^{n}})$ for small
$n$ and $p$ checked with computer algebra systems pari/gp and Sage.

  1. 1.

Let $n=3,p=5$ and $\alpha\in\mathbb{F}_{125}$ one of the roots of
$x^{3}+x^{2}+1$. Then $K_{1}(\alpha,\mathbb{F}_{125})=(3+\sqrt{5})/2$. On the
other hand if $A=\left[\begin{smallmatrix}0&1&0\\\ 0&0&1\\\
-1&0&-1\end{smallmatrix}\right]$ then $K_{3}(A,\mathbb{F}_{5})=327.2542$ which
agrees with $125K_{1}(\alpha,\mathbb{F}_{125})$.

  2. 2.

Let $n=4$ and $p=3$, $\alpha\in\mathbb{F}_{81}$ be a root of $x^{4}+2x^{3}+2$,
and $A=\left[\begin{smallmatrix}0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\
1&0&0&1\end{smallmatrix}\right]$. Then $K_{4}(A,\mathbb{F}_{3})=11664$ which
agrees with $-729K_{1}(\alpha,\mathbb{F}_{81})$.

  3. 3.

Let $n=3$, $p\equiv 1\ (3)$ and $A=\left[\begin{smallmatrix}0&1&0\\\ 0&0&1\\\
\mu&0&0\end{smallmatrix}\right]$, where
$\mu\in\mathbb{F}_{p}^{*}\setminus(\mathbb{F}_{p}^{*})^{3}$. If
$\alpha\in\mathbb{F}_{p^{3}}$ is such that $\alpha^{3}=\mu$ then one can check
that

$K(\alpha)=\sum e(3\mu c+(3a^{2}-3\mu cb)/\Delta(a,b,c)),$

where $e(x)=e^{2\pi ix}$, $\Delta(a,b,c)=a^{3}-3\mu cba+(\mu
b^{3}+\mu^{2}c^{3})$ and where the sum is over
$(a,b,c)\in\mathbb{F}_{p}^{3}\setminus\\{(0,0,0)\\}$.

A direct calculation using Bruhat decomposition shows that the conjecture in
this case is equivalent to

$K(\alpha)=K_{13}(A)+(1+({\mu}/{p}))q,$

where $({\mu}/{p})$ is the Legendre symbol and where

$K_{13}(A)=\sum_{t_{1},t_{2},t_{3}\in\mathbb{F}_{p}^{*}}e(\mu
t_{1}^{2}t_{3}^{2}+t_{1}^{2}t_{3}+\mu t_{3}+1/t_{2}-1/(\mu t_{1}t_{3}^{2})).$

Up to about $p\leq 200$, this can be checked fast even on a personal computer.
For example the order of $2\mod 199$ is $99$, and we have that

$K(\sqrt[3]{2},\mathbb{F}_{199^{3}})=3869.8269$

while for $A=\left(\begin{smallmatrix}0&1&0\\\ 0&0&1\\\
2&0&0\end{smallmatrix}\right)$

$K_{13}(A)=4267.8269$

with a difference of 398, which shows that
$K_{3}(A,\mathbb{F}_{199})=K(\sqrt[3]{2},\mathbb{F}_{199^{3}})$. On the other
hand $3\mod 199$ is a primitive root, and we have that for
$A=\left(\begin{smallmatrix}0&1&0\\\ 0&0&1\\\ 3&0&0\end{smallmatrix}\right)$

$K(\sqrt[3]{3},\mathbb{F}_{199^{3}})=K_{13}(A)=-2875.1994.$

We also checked all $p=3k+1\leq 200$, for which $2$ is not a cube mod $p$ and
found that $K(\sqrt[3]{2},\mathbb{F}_{p^{3}})=K_{3}(A,\mathbb{F}_{p})$ holds
for all of them.

######  Proposition 5.12.

Assume that $q$ is odd and let
$\alpha\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$ and
$a=\left(\begin{smallmatrix}0&1\\\
-N(\alpha)&-\operatorname{Tr}(\alpha)\end{smallmatrix}\right)$, where $N$ and
$\operatorname{Tr}$ are the norm and trace of the field extension
$\mathbb{F}_{q^{2}}|\mathbb{F}_{q}$. Then

$K_{2}(a,M_{2}(\mathbb{F}_{q}))=-qK_{1}(\alpha,\mathbb{F}_{q^{2}}^{*}).$

######  Remark 9.

The cohomology complex $H^{\bullet}_{c}$ corresponding to the sum
$K_{2}(a,M_{2}(\mathbb{F}_{q}))$ satisfies

$\dim H^{i}_{c}=\begin{cases}0,&\mathrm{if~{}}i\neq 4\\\
4,&\mathrm{if~{}}i=4\end{cases}$

and the Frobenius eigenvalues $\mu_{i}$ on $H^{4}_{c}$ satisfy
$\mu_{1}^{2}=q^{2}\lambda_{1}^{2},\mu_{2}^{2}=q^{4},\mu_{3}^{2}=q^{4}$, and
$\mu_{4}^{2}=q^{2}\lambda_{2}^{2}$ where $\lambda_{2}=\overline{\lambda}_{1}$
are the eigenvalues corresponding to $K_{1}(\alpha,\mathbb{F}_{q^{2}}^{*})$.
Apart from permutations the proposition determines the sign of the square
roots, we have $\mu_{1}=-q\lambda_{1}$, $\mu_{2}=q^{2}$, $\mu_{3}=-q^{2}$ and
$\mu_{4}=-q\lambda_{2}$. Thus
$K_{2}(a,M_{2}(\mathbb{F}_{q^{m}}))=\sum_{i=1}^{4}\mu_{i}^{m}.$ Note that here
again we have cancellation: $\mu_{2}^{m}+\mu_{3}^{m}=0\iff 2\nmid m$.

###### Proof of Proposition 5.12.

As above let $K=\mathbb{F}_{q}[a]$ be the subring of $M_{2}$ generated by $a$.
$K$ is isomorphic $\mathbb{F}_{q^{2}}$ by the assumption on $\alpha$. Also,
the vectorspace $\mathbb{F}_{q}^{2}$ as an $\mathbb{F}_{q}[a]$-module is
isomorphic to the $\mathbb{F}_{q}$ vector space $\mathbb{F}_{q^{2}}$, with $a$
acting via multiplication by $\alpha$. For the moment denote this
$L_{\alpha}$, $L_{\alpha}:\beta\mapsto\alpha\beta$. Let
$\mathbb{F}_{q^{2}}\langle\tau\rangle$ be the non-commutative ring of twisted
polynomials, $\sum_{i}\xi_{i}\tau^{i}$ subject to $\tau\xi=F(\xi)\tau$, where
$F(\xi)=\xi^{q}$ is the Frobenius automorphism of
$\mathbb{F}_{q^{2}}/\mathbb{F}_{q}$.

There is an obvious map from
$\mathcal{M}_{2}=\mathbb{F}_{q^{2}}\langle\tau\rangle/(\tau^{2}-1)$ to
$M_{2}(\mathbb{F}_{q})$, sending $\xi_{1}+\xi_{2}\tau$ to the
$\mathbb{F}_{q}$-linear transformation $L_{\xi_{0}}+L_{\xi_{1}}F$. It is not
difficult to see that this linear map is injective, and so by dimension count,
an isomorphism. This identifies $M_{2}$ with $\mathcal{M}_{2}$, and it is easy
to check that under this identification
$\psi(L_{\xi_{0}}+L_{\xi_{1}}F)=\varphi_{2}(\xi_{0})$, where
$\varphi_{2}=\varphi\circ\operatorname{Tr}_{\mathbb{F}_{q^{2}}/\mathbb{F}_{q}}$.
It follows that

$\sum_{x\in
M_{2}(\mathbb{F}_{q})^{*}}\psi(ax+x^{-1})=\sum_{\xi_{0}+\xi_{1}\tau\in\mathcal{M}_{2}^{*}}\varphi_{2}(\alpha\xi_{0}+(\xi_{0}+\xi_{1}\tau)^{-1}).$

An easy calculations shows that $(1+\xi\tau)\in\mathcal{M}_{2}^{*}$ exactly
when $N(\xi)\neq 1$, and then
$(1+\xi\tau)^{-1}=\frac{1}{1-N(\xi)}(1-\xi\tau)$. One also has that
$(\xi\tau)^{-1}=F(\xi^{-1})\tau$ and so

$\mathcal{M}_{2}^{*}=\\{\xi_{1}\tau\,|\,\xi_{1}\in\mathbb{F}_{q^{2}}^{*}\\}\cup\\{\xi_{0}(1+\xi_{1}\tau)\,|\,\xi_{0}\in\mathbb{F}_{q^{2}}^{*},\xi_{1}\in\mathbb{F}_{q^{2}},,N(\xi_{1})\neq
1\\}.$

Therefore

$\sum_{x\in\mathrm{GL}_{2}(\mathbb{F}_{q})}\psi(ax+x^{-1})=q^{2}-1+\sum_{\begin{subarray}{c}\xi_{0}\in\mathbb{F}_{q^{2}}^{*},\xi_{1}\in\mathbb{F}_{q^{2}}\\\
N(\xi_{1})\neq
1\end{subarray}}\varphi_{2}\left(\alpha{\xi_{0}}+(1-N(\xi_{1}))^{-1}\xi_{0}^{-1}\right).$

Now the norm map $N$ is a surjective homomorphism from
$\mathbb{F}_{q^{2}}^{*}\to\mathbb{F}_{q}^{*}$, with a kernel of size $q+1$ and
so for $\gamma\in\mathbb{F}_{q}$

$|\\{\xi\in\mathbb{F}_{q^{2}}\,|\,(1-N(\xi))^{-1}=\gamma\\}|=\begin{cases}0&\text{
if }\gamma=0\\\ 1&\text{ if }\gamma=1\\\ q+1&\text{ if }\gamma\neq
0,1.\end{cases}$

This gives

$\sum_{x\in\mathrm{GL}_{2}(\mathbb{F}_{q})}\psi(ax+x^{-1})=\\\
q^{2}-1+(q+1)\sum_{\begin{subarray}{c}\xi_{0}\in\mathbb{F}_{q^{2}}^{*}\\\
\gamma\in\mathbb{F}_{q}^{*}\end{subarray}}\phi^{(2)}\left(\alpha{\xi_{0}}+\gamma\xi_{0}^{-1}\right)-q\sum_{\xi_{0}\in\mathbb{F}_{q^{2}}^{*}}\phi^{(2)}\left(\alpha{\xi_{0}}+\xi_{0}^{-1}\right).$

Finally

$\sum_{\begin{subarray}{c}\xi_{0}\in\mathbb{F}_{q^{2}}^{*}\\\
\gamma\in\mathbb{F}_{q}^{*}\end{subarray}}\phi^{(2)}\left(\alpha{\xi_{0}}+\gamma\xi_{0}^{-1}\right)=\sum_{\begin{subarray}{c}\xi_{0}\in\mathbb{F}_{q^{2}}^{*}\\\
\gamma\in\mathbb{F}_{q}^{*}\end{subarray}}\phi^{(2)}\left(\alpha{\xi_{0}}\right)\phi\left(\gamma\operatorname{Tr}\xi_{0}^{-1}\right)\\\
=-\sum_{\begin{subarray}{c}\xi_{0}\in\mathbb{F}_{q^{2}}^{*}\\\
\operatorname{Tr}\xi_{0}^{-1}=0\end{subarray}}\phi^{(2)}\left(\alpha{\xi_{0}}\right)=-(q-1)$

∎

###  5.5 The purity locus

We have seen that for a regular semi-simple element $a\in
M_{n}(\mathbb{F}_{q})$ the Kloosterman sum $K_{n}(a)$ is pure. The tables
above already suggest that for a matrix $a$ with more than one Jordan block
for an eigenvalue, that sum can not be pure. This can be seen without
reference to cohomology. To see this assume that $a$ has a single eigenvalue
$\alpha$. Recall that

$K_{n}(a)=P(A,G,K)=\sum_{2f\leq n}c_{f}(q,q-1)K^{n-2f},$

where $c_{f}$ are polynomials. We give the $A$ and $G$ weigth 1, and $K$
weight $1/2$, so the polynomial $P$ has a weighted degree, which determines
the order of magnitude (in $q$) of its value . Now $f=1$ corresponds to simple
transpositions, and by Theorem 5.8 one sees that these sums are too large in
magnitude to be pure if not all $\varepsilon_{i}$ are 0.

It is an intriguing question what happens for $K_{[n]}$ when $a$ has only one
Jordan block. The recursion formula gives

$K_{[n]}=q^{n-1}KK_{[n-1]}-q^{2n-2}K_{[n-2]}$

where $K=K_{1}(\alpha)$. Let $k_{n}=q^{-n(n-1)/2}K_{[n]}$, so that we have

$k_{n}=Kk_{n-1}-qk_{n-2}.$

It follows that there exist $c_{1},c_{2}$ such that

$k_{n}=c_{1}\lambda_{1}^{n}+c_{2}\lambda_{2}^{n}$

where $\lambda_{1},\lambda_{2}$ are the roots of $X^{2}-KX+q$. These are
exactly the eigenvalues of Frobenius acting on the cohomology of the
Kloosterman sheaf. Using that $K=\lambda_{1}+\lambda_{2}$, and that
$K_{[2]}=-q^{2}+K_{1}^{2}q$ we get that $k_{1}=\lambda_{1}+\lambda_{2}$, and
that
$k_{2}=k_{1}^{2}-q=\lambda_{1}^{2}+\lambda_{1}\lambda_{2}+\lambda_{2}^{2}$.
Therefore $c_{1}=\frac{\lambda_{1}}{\lambda_{1}-\lambda_{2}}$,
$c_{2}=-\frac{\lambda_{2}}{\lambda_{1}-\lambda_{2}}$ and

$k_{n}=\frac{\lambda_{1}^{n+1}-\lambda_{2}^{n+1}}{\lambda_{1}-\lambda_{2}}=\sum_{j=0}^{n}\lambda_{1}^{j}\lambda_{2}^{n-j}.$
(53)

This evaluation has an interesting interpretation. Let
$X^{2}-K_{1}X+q=(X-\lambda_{1})(X-\lambda_{2})$, with
$\lambda_{1,2}=q^{1/2}e^{\pm i\theta}$, so that $K_{1}=2q^{1/2}\cos\theta$. We
have that

$k_{n}=\frac{\lambda_{1}^{n+1}-\lambda_{2}^{n+1}}{\lambda_{1}-\lambda_{2}}=q^{n/2}\frac{e^{(n+1)\theta}-e^{-(n+1)\theta}}{e^{\theta}-e^{-\theta}}=\frac{\sin(n+1)\theta}{\sin\theta}$

Therefore

$K_{[n]}=q^{n(n-1)/2}U_{n}(\cos\theta)$ (54)

where $U_{n}$ is the Chebyshev polynomial of the second kind. The Sato-Tate
distribution of the angles of $K_{1}(\alpha)$ over the valuations of a global
field is then equivalent to non-trivial cancellation in the sums

$\sum_{N(v)\leq x}K_{n}(a,\mathbb{F}_{v})/N(v)^{n(n-1)/2},$

where $a=\alpha{I}+\sum_{i=1}^{n-1}\mathbf{e}_{i,i+1}$.

Getting back to the question of purity these sums are pure from a numerical
point of view, but this in itself does not rule out a cohomology with a
nilpotent Frobenius action.

For example in the case $n=2$ it is easy to see that the cohomologies
corresponding to the Bruhat cells are as follows.

On $C_{I}=B$ the trace of $\alpha x+x^{-1}$ can be written as a product of two
Kloosterman sums over the diagonal elements, thus we have

$H^{\bullet}_{C_{I}}=H^{\bullet}(C_{I},x\mapsto
tr(ax+x^{-1}))=H^{\bullet}_{c}(\mathbb{A}^{1}-\mathbb{A}^{0},f_{\alpha})\otimes
H^{\bullet}_{c}(\mathbb{A}^{1}-\mathbb{A}^{0},f_{\alpha})\otimes
H^{\bullet}_{c}(\mathbb{A}^{1},0).$

That implies $\dim(H^{i}_{C_{I}}=0$ unless $i=4$ and $\dim(H^{4}_{C_{I}})=4$.

On the nontrivial cell $C_{w}=UwB$ we have seen that the sum (and the
cohomology) cancels on the subvariety $\alpha\det(b)\neq 1$ and on the rest we
have

$H^{\bullet}_{C_{w}}=H^{\bullet}(C_{w},x\mapsto\operatorname{tr}(ax+x^{-1}))=H^{\bullet}_{c}(\mathbb{A}^{1}-\mathbb{A}^{0},\mathrm{id})\otimes
H^{\bullet}_{c}(\mathbb{A}^{1},0)\otimes H^{\bullet}_{c}(\mathbb{A}^{1},0).$

That implies $\dim(H^{i}_{C_{w}})=0$ unless $i=5$ and $\dim(H^{5}_{C_{w}})=1$.

Thus the long exact sequence of the excision ($C_{I}=G\setminus C_{w}$) gives

$0\to H^{4}_{G}\to H^{4}_{C_{I}}\to H^{5}_{C_{w}}\to H^{5}_{G}\to 0.$

Either $\dim(H^{4}_{G})=4$ and $\dim(H^{5}_{G})=1$ or $\dim(H^{4}_{G})=3$ and
$\dim(H^{5}_{G})=0$ seems to be possible.

The same problem exists for higher degree cases333While the paper was reviewed
for publication the purity of these sums was proved in [EST]..

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