$\displaystyle=(x_{1},\cdots,x_{m})\begin{pmatrix}1&0&&\cdots&0\\\
h_{1,2}&1&0&\cdots&0\\\ h_{1,3}&h_{2,3}&1&&0\\\
\vdots&&\ddots&\ddots&\vdots\\\
h_{1,m}&h_{2,m}&\cdots&h_{m-1,m}&1\end{pmatrix}$
$\displaystyle=(\sum_{k=1}^{m}x_{k}h_{1,k},\cdots,x_{m-1}+x_{m}h_{m-1,m},x_{m})$

So

$\mathbf{x}H^{T}-\mathbf{x}=(\sum_{k=2}^{m}x_{k}h_{1,k},\cdots,x_{m}h_{m-1,m},0).$

Thus

$\displaystyle\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})=\sum_{k=1}^{m-1}a_{k}\cdot(\sum_{j=k+1}^{m}x_{j}h_{k,j})$

######  Calculation 1.

For $j_{0}>1$, $\mathbf{x}=(x_{i})$ with $x_{i}=0$ except for $x_{j_{0}}$.

Then we get

$\displaystyle\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})=\sum_{k=1}^{j_{0}-1}a_{k}\cdot
x_{j_{0}}h_{k,j_{0}}=x_{j_{0}}\left(\sum_{k=1}^{j_{0}-1}a_{k}h_{k,j_{0}}\right)$

So for all $j_{0}>2$, we must have

$\sum_{k=1}^{j_{0}-1}a_{k}h_{k,j_{0}}=0.$

So if $a_{1}\neq 0$, given $h_{i,j_{0}}$ for $i\neq 1,k$, the above dictates
$h_{1,j_{0}}$:

$h_{1,j_{0}}=\frac{-1}{a_{1}}\sum_{k=2}^{j_{0}-1}a_{k}h_{k,j_{0}}.$

Therefore,

$|L_{s}|\leq|\\{H:H_{1,j}\text{ fixed }\forall j\neq
1\\}=|\text{Up}_{n-1}(\mathbb{F}_{p^{r}})|=p^{r(n-1)(n-2)/2}.$

If $a_{i}=0$ for $i\neq 0$, then we get from the calculation above that

$h_{1,j_{0}}=\frac{-1}{a_{1}}\sum_{k=2}^{j_{0}-1}a_{k}h_{k,j_{0}}=0.$

So

$\displaystyle\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})$
$\displaystyle=\sum_{k=1}^{m-1}a_{k}\cdot(\sum_{j=k+1}^{m}x_{j}h_{k,j})$
$\displaystyle=a_{1}\cdot(\sum_{j=2}^{m}x_{j}h_{1,j}),$ $\displaystyle\text{
since }a_{i}=0,i>1$ $\displaystyle=0$ $\displaystyle\text{ since
}h_{1,j}=0,j>1$

So we get that for $s=(a_{1},0,\cdots,0)$ with $a_{1}\neq 0$,

$L_{s}=\\{H:H_{1,j}=0,\forall j\neq 1\\}.$

∎

###  6.5 Proof

###### Proof of Theorem 6.1.

######  Case 1 ($n=4$).

###### Proof of Theorem 6.1 for $n=2$.

For $p\neq 2$, the action of
$\text{Up}_{2}(\mathbb{F}_{p^{r}})\cong\mathbb{F}_{p^{r}}$ on
$\text{Antisym}(2,p^{r})\cong\mathbb{F}_{p^{r}}$ is trivial and so
$\text{Syl}_{p}(\Omega^{+}(2n,p^{r}))\cong\mathbb{F}_{p^{r}}\times\mathbb{F}_{p^{r}}$.
So
$\text{ed}_{k}(\text{Syl}_{p}(\Omega^{+}(4,p^{r})))=\text{ed}_{k}(\mathbb{F}_{p^{r}}\times\mathbb{F}_{p^{r}})=2\text{ed}_{k}(\mathbb{F}_{p^{r}})=2r$.
This proves Theorem 6.1 for $n=4$.

Similarly for $n=4$, $p=2$, the action of
$\text{Up}_{2}(\mathbb{F}_{p^{r}})\cong\mathbb{F}_{2^{r}}$ on
$\text{Antisym0}(2,2^{r})\cong\mathbb{F}_{2^{r}}$ is trivial and so
$\text{Syl}_{2}(\Omega^{+}(2n,2^{r}))\cong\mathbb{F}_{2^{r}}\times\mathbb{F}_{2^{r}}$.
So
$\text{ed}_{k}(\text{Syl}_{2}(\Omega^{+}(4,2^{r})))=\text{ed}_{k}(\mathbb{F}_{2^{r}}\times\mathbb{F}_{2^{r}})=2\text{ed}_{k}(\mathbb{F}_{2^{r}})=2r$.
This proves Theorem 6.1 for $n=4$.

∎

######  Remark 31.

The work in the previous section is valid, though unnecessary, for $n=4$. It
gives us that the minimum dimension of an irrep is $1$. And then since the
center has rank $2r$, we will get an essential dimension of $2r$.

######  Remark 32.

Alternate proof:

$\displaystyle\Omega^{+}(4,p^{r})\cong(SL_{2}(\mathbb{F}_{p^{r}})\times
SL_{2}(\mathbb{F}_{p^{r}}))/\pm 1$
$\displaystyle\Rightarrow\text{ed}_{k}(\Omega^{+}(4,p^{r}),p)=\text{ed}_{k}(SL_{2}(\mathbb{F}_{p^{r}}),p)+\text{ed}_{p}(SL_{2}(\mathbb{F}_{p^{r}}),p)=2r$

.

######  Case 2 ($n=5$).

Step 1: Find the lower bound

Since $\\{\chi_{i}|_{Z(\text{Syl}_{p}(\Omega(5,p^{r})))}\\}$ is a basis for
$\widehat{Z(\text{Syl}_{p}(\Omega(5,p^{r})))}$. We can conclude that for at
least $r$ of the $\chi_{i}=\psi_{(b_{i})}$, we have $b_{1}\neq 0$ and so the
dimension is $\geq p^{r}$. The others might have $b_{1}=0$ and so could have
dimension 1. Thus we must have that $\dim(\rho)\geq rp^{r}+r$. So

$\text{ed}_{k}(\text{Syl}_{p}(\Omega(5,p^{r})))\geq rp^{r}+r.$

Step 2: Construct the upper bound

Let $e_{i}$ be a basis for $\mathbb{F}_{p^{r}}^{+}$ over
$\mathbb{Z}/p\mathbb{Z}$. Then $T=\\{(e_{i},0),(0,e_{i})\\}$ forms a basis for
$(\mathbb{F}_{p^{r}}^{+})^{2}$ over $\mathbb{Z}/p\mathbb{Z}$. Then by Lemma 7,
$\\{\psi_{(e_{i},0)},\psi_{(0,e_{i})}\\}$ forms a basis for the characters of
$(\mathbb{F}_{p^{r}}^{+})^{2}=\Omega_{1}((\mathbb{F}_{p^{r}})^{2})$. Let

$S=\\{(e_{i},0,0),(0,0,e_{i})\\}.$

Then for $s\in S$, $\theta_{s,\lambda}$ have central character $\psi_{t}$
where $t=(s_{1},s_{3})$. By Corollary 6.33, for $\lambda$ trivial,

$\dim(\theta_{s,\lambda})=\begin{cases}p^{r},&\text{ if }s=(e_{i},0,0)\\\
1,&\text{ if }s=(0,0,e_{i})\end{cases}.$

Let $\rho=\oplus_{s\in S}\theta_{s,\lambda}$. Then

$\displaystyle\dim(\rho)=\sum_{s\in S}\dim(\theta_{s,\lambda})=rp^{r}+r$

And by Lemma 3.3, since $\\{\psi_{(e_{i},0)},\psi_{(0,e_{i})}\\}$ forms a
basis for the characters of
$(\mathbb{F}_{p^{r}}^{+})^{2}=\Omega_{1}((\mathbb{F}_{p^{r}})^{2})$, we have
that $\rho=\oplus_{s\in S}\theta_{s,\lambda}$ is a faithful representation of
$\text{Syl}_{p}(\Omega(5,p^{r}))$. Thus $\rho$ is the desired faithful
representation of $\text{Syl}_{p}(\Omega(5,p^{r}))$ of dimension $rp^{r}+r$.

######  Case 3 ($n=2m,m>2$).

Step 1: Find the lower bound

The minimum dimension of an irreducible representation of
$\text{Syl}_{p}(\Omega^{+}(2n,p^{r})$ with non-trivial central character is
given by $\theta_{s,\lambda}$ with $\lambda$ trivial and $s=(b,0,\cdots,0)$.
So the minimum dimension is $p^{2r(m-2)}$. Thus the minimum dimension of a
faithful representation $rp^{2r(m-2)}$ since the center has rank $r$.

Step 2: Construct the upper bound

Let $\\{e_{i}\\}$ be a basis for $\mathbb{F}_{p^{r}}^{+}$ over
$\mathbb{Z}/p\mathbb{Z}$. Then by Lemma 7, $\\{\psi_{e_{i}}\\}$ form a basis
for the characters of $\mathbb{F}_{p^{r}}^{+}$. Let
$s_{i}=(e_{i},0,\cdots,0)$. Then by Corollary 6.30, for $\lambda$ trivial,

$\dim(\theta_{s_{i},\lambda})=p^{2r(n-2)}.$

Let $\rho=\oplus_{i}\theta_{s_{i},\lambda}$. Then

$\dim(\rho)=\sum_{i=1}^{r}\dim(\theta_{s_{i},\lambda})=\sum_{i=1}^{r}p^{r(m-2)}=rp^{2r(m-2)}.$

And by Lemma 3.3, since $\psi_{e_{i}}|_{\mathbb{F}_{p^{r}}}$ form a basis for
$\widehat{\mathbb{F}_{p^{r}}}$, we have that
$\oplus_{i}\theta_{s_{i},\lambda}$ is a faithful representation of
$\text{Syl}_{p}(\Omega^{+}(2m,p^{r}))$ Thus $\rho$ is the desired faithful
representation of $\text{Syl}_{p}(\Omega^{+}(2m,p^{r}))$ of dimension
$rp^{2r(m-2)}.$

######  Case 4 ($n=2m+1,m>2$).

Step 1: Find the lower bound

Since $\\{\chi_{i}|_{Z(\text{Syl}_{p}(\Omega(2m+1,p^{r})))}\\}$ a basis for
$\widehat{Z(\text{Syl}_{p}(\Omega(2m+1,p^{r})))}$, we must have $b_{1}\neq 0$
for at least $r$ of the $\rho_{i}$. So the minimum dimension is
$p^{r(m-1)(m-2)}$. The other $r$ may have $b_{1}=0$ but then we must have
$a_{1}\neq 0$, so the minimum dimension is $p^{r(m-1)}$. Thus we have

$\text{ed}_{k}\left(\text{Syl}_{p}(\Omega(2m+1,p^{r}))\right)\geq
rp^{r(m-1)(m-2)}+rp^{r(m-1)}=rp^{r(m-1)}(p^{r(m-2)}+1).$

Step 2: Construct the upper bound

Let $\\{e_{i}\\}$ be a basis for $\mathbb{F}_{p^{r}}^{+}$ over
$\mathbb{Z}/p\mathbb{Z}$. Then by Lemma 7, $\\{\psi_{e_{i}}\\}$ form a basis
for the characters of $\mathbb{F}_{p^{r}}^{+}$. Let

$S=\\{(e_{i},0,\cdots,0),(0,\cdots,0,e_{i},0,\cdots,0)\\}.$

For $s\in S$, $\\{\psi_{s}|_{Z(\text{Syl}_{p}(\Omega(2m+1,p^{r}))}\\}$ forms a
basis for $\widehat{Z(\text{Syl}_{p}(\Omega(2m+1,p^{r})))}$. Let

$\rho=\oplus_{s\in S}\theta_{s,\lambda}.$

Then by Lemma 3.3, $\rho=\oplus_{s\in S}\theta_{s,\lambda}$ is a faithful
representation of $\text{Syl}_{p}(\Omega(2m+1,p^{r}))$. By Corollaries 6.36
and 6.38, for $s\in S$, $\lambda$ trivial,

$\dim(\theta_{s,\lambda})=\begin{cases}p^{r(m-1)},&s=(e_{i},0,\cdots,0)\\\
p^{r(m-1)(m-2)},&s=(0,\cdots,0,e_{i},0,\cdots,0)\end{cases}.$

Thus

$\dim(\rho)=\sum_{s\in
S}\dim(\theta_{s,\lambda})=rp^{r(m-1)}+rp^{r(m-1)(m-2)}=rp^{r(m-1)}(p^{r(m-2)}+1).$

Thus $\rho$ is the faithful representation of
$\text{Syl}_{p}(\Omega(2m+1,p^{r}))$ of desired dimension. ∎

##  7 The Unitary Groups

In this section, we will show the following theorem:

######  Theorem 7.1.

For any $p$, $k$ a field such that $\text{char }k\neq p$,

$\text{ed}_{k}(PSU(n,p^{2r}),p)=\text{ed}_{k}(U(n,p^{2r}),p)=\begin{cases}rp^{2r(m-1)}&n=2m\\\
3rp^{2r(m-1)}&n=2m+1\end{cases}$

###  7.1 Definitions

For any $p$, by the Artin-Schreier Theorem ([Lan02], Theorem 6.4) we can write
$\mathbb{F}_{p^{2r}}=\mathbb{F}_{p^{r}}[\alpha]$ where $\alpha$ is a root of
$x^{2}-x-\eta$. Then
$\mathbb{F}_{p^{2r}}=\mathbb{F}_{p^{r}}+\alpha\mathbb{F}_{p^{r}}$.

######  Definition 7.2.

For $p\neq 2$, we can change $x^{2}-x-\eta$ by a linear change of variables to
the form $x^{2}+\eta^{\prime}$. Let $\alpha$ be a root of
$x^{2}+\eta^{\prime}$. Then we still have
$\mathbb{F}_{p^{2r}}=\mathbb{F}_{p^{r}}+\alpha\mathbb{F}_{p^{r}}$. We will
write elements in $\mathbb{F}_{p^{2r}}$ as $a=\dot{a}+\ddot{a}\alpha$, where
$\dot{a},\ddot{a}\in\mathbb{F}_{p^{r}}$. For $a=\dot{a}+\ddot{a}\alpha$,
define $\overline{a}=\dot{a}-\ddot{a}\alpha$.

######  Definition 7.3.

For $p=2$, let $\alpha$ be a root of $x^{2}+x+\eta$. Note that
$(\alpha+1)^{2}+(\alpha+1)+\eta=\alpha^{2}+\alpha+\eta=0.$ So $\alpha+1$ gives
the other root of $x^{2}+x+\eta$. We will write elements in
$\mathbb{F}_{p^{2r}}$ as $a=\dot{a}+\ddot{a}\alpha$, where
$\dot{a},\ddot{a}\in\mathbb{F}_{p^{r}}$. For $a=\dot{a}+\ddot{a}\alpha$,
define
$\overline{a}=\dot{a}+\ddot{a}(\alpha+1)=(\dot{a}+\ddot{a})+\ddot{a}\alpha$.

######  Remark 33.

For any $p$, we have that for all $a\in\mathbb{F}_{p^{2r}}$,
$\overline{\overline{a}}=a$.

######  Definition 7.4.

For $\mathbf{x}=(x_{i})\in(\mathbb{F}_{p^{2r}})^{n}$, let
$\overline{\mathbf{x}}=(\overline{x_{i}})$.

######  Definition 7.5.

For $M=(M_{ij})\in GL(n,\mathbb{F}_{p^{2r}})$, let
$\overline{M}=(\overline{M_{ij}})$.

######  Definition 7.6.

A matrix $\beta$ is called Hermitian if $\beta^{T}=\overline{\beta}$.

######  Definition 7.7.

Given a Hermitian matrix $\beta$, define

$U_{\beta}(n,p^{2r})=\\{M\in
GL(n,\mathbb{F}_{p^{2r}}):M^{T}\beta\overline{M}=\beta\\}.$

By Grove ([Gro00], Corollary 10.4), for any $p,r,n,\beta,\beta^{\prime}$,
$U_{\beta}(n,p^{2r})\cong U_{\beta^{\prime}}(n,p^{2r})$, so we may work with
whichever choice of $\beta$ we like. So for the remainder of this paper, we
will drop the $\beta$ in our notation.

######  Definition 7.8.

Define $SU(n,p^{2r})$ as $SU(n,p^{2r})=\\{M\in U(n,p^{2r}):\det(M)=1\\}$.

######  Definition 7.9.

Define $PSU(n,p^{2r})$ as $PSU(n,p^{2r})=SU(n,p^{2r})/(Z(SU(n,p^{2r}))$.

######  Definition 7.10.

Just as for the Orthogonal groups the Witt index is defined to be the
dimension of a maximal totally isotropic subspace of $\mathbb{F}_{p^{r}}$. Let
$w$ denote the Witt index.

######  Remark 34.

Grove shows ([Gro00], Theorem 11.26) that for $n\geq 2$ and Witt index $w\geq
1$, $PSU(n,p^{r})$ is simple except for $(n,p^{r})=(2,4),(2,9),$ or $(3,4)$.
Grove also shows ([Gro00], Propoisition 11.17) that

$Z(SU(n,p^{2r}))=\\{a\text{Id}_{n}:a\in\mathbb{F}_{p^{2r}},a\overline{a}=1,\text{
and }a^{n}=1\\}.$

###  7.2 The $p$-Sylows

By Grove ([Gro00], Theorem 11.28 and Cor 11.29),
$|\text{Syl}_{p}(U(n,p^{2r}))|=p^{rn(n-1)/2}=|\text{Syl}_{p}(PSU(n,p^{2r}))|.$
So $\text{Syl}_{p}(U(n,p^{r}))$ maps injectively onto
$\text{Syl}_{p}(PSU(n,p^{r}))$; thus their $p$-Sylows must be isomorphic, and
hence they have the same essential $p$-dimension.

####  7.2.1 $n=2m$

If $n=2m$, then we can choose

$\beta=\begin{pmatrix}0_{m}&\text{Id}_{m}\\\
\text{Id}_{m}&0_{m}\end{pmatrix}.$

######  Remark 35.

$M=\begin{pmatrix}A&B\\\ C&D\end{pmatrix},\in
U_{\beta}(n,p^{2r})\Leftrightarrow
A^{T}\overline{C}+C^{T}\overline{A}=0=B^{T}\overline{D}+D^{T}\overline{B}\text{
and
}A^{T}\overline{D}+C^{T}\overline{B}=\text{Id}_{m}=B^{T}\overline{C}+D^{T}\overline{A}.$

######  Definition 7.11.

For any $p$, define

$G_{p^{r}}=\\{a\in\mathbb{F}_{p^{2r}}:\overline{a}=-a\\}.$

######  Remark 36.

$G_{p^{r}}=\begin{cases}\\{\dot{a}+\ddot{a}\alpha:\dot{a}=0\\}=\alpha\mathbb{F}_{p^{r}}\cong(\mathbb{Z}/p\mathbb{Z})^{r}&p\neq
2\\\
\\{\dot{a}+\ddot{a}\alpha:\ddot{a}=0\\}=\mathbb{F}_{p^{r}}\cong(\mathbb{Z}/p\mathbb{Z})^{r}&p=2\end{cases}$

######  Definition 7.12.

Let $\text{antisym*}(m,p^{2r})=\\{B\in GL(n,p^{2r}):B^{T}=-\overline{B}\\}$

To construct a matrix in $\text{antisym*}(m,p^{2r})$, we can choose the
entries in the upper right to be anything, and this will dictate the entries
in the lower right. On the diagonal, we must have
$B_{i,i}=-\overline{B_{i,i}}$. So

$|\text{antisym*}(m,p^{2r})|=|\text{Up}_{m}(\mathbb{F}_{p^{2r}})|\cdot|G_{p^{r}}|^{m}=(p^{2r})^{\frac{m(m-1)}{2}}\cdot(p^{r})^{m}=p^{rm(m-1)}\cdot
p^{rm}=p^{rm^{2}-rm+rm}=p^{rm^{2}}$

######  Proposition 7.13.

For $n=2m$, any $p$,

$\text{Syl}_{p}(U(n,p^{2r}))=\\{\begin{pmatrix}A&0_{m}\\\
0_{m}&(\overline{A^{-1}})^{T}\end{pmatrix}\begin{pmatrix}\text{Id}_{m}&B\\\
0_{m}&\text{Id}_{m}\end{pmatrix}:A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}}),B\in\text{antisym*}(m,p^{2r})\\}.$

###### Proof.

Let

$O=\\{\begin{pmatrix}A&0_{m}\\\
0_{m}&(\overline{A^{-1}})^{T}\end{pmatrix}:A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}})\\}.$

This is a subgroup of $U(2m,p^{2r})$ isomorphic to
$\text{Up}_{m}(\mathbb{F}_{p^{2r}})$. So
$|O|=|\text{Up}_{m}(\mathbb{F}_{p^{2r}})|=p^{rm(m-1)}.$ Let

$N=\\{\begin{pmatrix}\text{Id}_{n}&B\\\
0_{n}&\text{Id}_{n}\end{pmatrix}:B\in\text{antisym*}(m,p^{2r})\\}.$

This is a subgroup of $U(2m,p^{2r})$ isomorphic to
$\text{antisym*}(n,p^{2r})$. So $|N|=|\text{antisym*}(m,p^{2r})|=p^{rm^{2}}$.
A direct computation shows that $N\cdot O=N\rtimes O$ and

$|N\rtimes O|=p^{rn(n-1)/2}=|\text{Syl}_{p}(U(n,p^{2r}))|.$

Thus

$N\rtimes O=\\{\begin{pmatrix}A&0_{n}\\\
0_{n}&(\overline{A^{-1}})^{T}\end{pmatrix}\begin{pmatrix}\text{Id}_{n}&B\\\
0_{n}&\text{Id}_{n}\end{pmatrix}:A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}}),B\in\text{antisym*}(n,p^{2r})\\}$

is a $p$-Sylow of $U(2m,p^{2r})$. ∎

######  Corollary 7.14.

For $n=2m$, any $p$,

$\text{Syl}_{p}(U(n,p^{2r}))\cong\text{antisym*}(m,p^{2r})\rtimes\text{Up}_{m}(\mathbb{F}_{p^{2r}}),$

where the action is given by

$A(B)=AB\overline{A}^{T}.$

####  7.2.2 $n=2m+1$

If $n=2m+1$, then we can choose

$\beta=\begin{pmatrix}-1&\mathbf{0}&\mathbf{0}\\\
\mathbf{0}&0_{m}&\text{Id}_{m}\\\ \mathbf{0}&\text{Id}_{m}&0_{m}\end{pmatrix}$

since $\overline{-1}=-1$ when $p$ is odd and $-1=1$ when $p$ is even.

######  Remark 37.

$M=\begin{pmatrix}a&\mathbf{w}&\mathbf{x}\\\ \mathbf{y}&A&B\\\
\mathbf{z}&C&D\end{pmatrix}$

is in $U(2m+1,p^{2r})$ if and only if

$\begin{pmatrix}A&B\\\ C&D\end{pmatrix}\in U(2m,p^{2r})$

and

$a\overline{a}-\mathbf{y}\cdot\overline{\mathbf{z}}-\mathbf{z}\cdot\overline{\mathbf{y}}=1$

and

$0=-a\overline{\mathbf{w}}+\mathbf{y}\overline{C}+\mathbf{z}\overline{A}=-a\overline{\mathbf{x}}+\mathbf{y}\overline{D}+\mathbf{z}\overline{B}=-a\overline{\mathbf{w}}+A^{T}\overline{\mathbf{z}}+C^{T}\overline{\mathbf{y}}=-a\overline{\mathbf{x}}+B^{T}\overline{\mathbf{z}}+D^{T}\overline{\mathbf{y}}$

If $A=D=\text{Id}_{m}$ and $B=C=0_{m}$, $a=1$, this becomes

$\mathbf{y}\cdot\overline{\mathbf{z}}=-\mathbf{z}\cdot\overline{\mathbf{y}},\text{
}0=-\overline{\mathbf{w}}+\mathbf{z}=-\overline{\mathbf{x}}+\mathbf{y}.$

Note this is satisfied when
$\mathbf{z}=\mathbf{w}=\mathbf{0},\mathbf{x}\in\mathbb{F}_{p^{2r}}^{m},\mathbf{y}=\overline{\mathbf{x}}$.

######  Proposition 7.15.

For $n=2m+1$, any $p$,

$\displaystyle\text{Syl}_{p}(U(n,p^{2r}))$
$\displaystyle=\\{\begin{pmatrix}1&\mathbf{0}&\mathbf{x}\\\
\overline{\mathbf{x}}^{T}&\text{Id}_{n}&0_{n}\\\
\mathbf{0}&0_{n}&\text{Id}_{n}\end{pmatrix}\begin{pmatrix}1&\mathbf{0}&\mathbf{0}\\\
\mathbf{0}&A&0_{n}\\\
\mathbf{0}&0_{n}&(\overline{A^{-1}})^{T}\end{pmatrix}\begin{pmatrix}1&\mathbf{0}&\mathbf{0}\\\
\mathbf{0}&\text{Id}_{n}&B\\\
\mathbf{0}&0_{n}&\text{Id}_{n}\end{pmatrix}:\mathbf{x}\in\mathbb{F}_{p^{2r}}^{m},A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}}),B\in\text{antisym*}(m,p^{2r})\\}.$

###### Proof.

Let

$P=\\{\begin{pmatrix}1&0&\mathbf{x}\\\
\overline{\mathbf{x}}^{T}&\text{Id}_{n}&0_{n}\\\
0&0_{n}&\text{Id}_{n}\end{pmatrix}:\mathbf{y}\in\mathbb{F}_{p^{2r}}^{m}\\}.$

This is a subgroup of $U(2n+1,p^{2r})$ isomorphic to
$(\mathbb{F}_{p^{2r}}^{+})^{m}$. So
$|P|=|(\mathbb{F}_{p^{2r}}^{+})^{m}|=p^{2rm}.$ Let

$N=\\{\begin{pmatrix}1&\mathbf{0}&\mathbf{0}\\\ \mathbf{0}&\text{Id}_{n}&B\\\
\mathbf{0}&0_{n}&\text{Id}_{n}\end{pmatrix}:B\in\text{antisym*}(m,p^{2r})\\}.$

This is a subgroup of $U(2n+1,p^{2r})$ isomorphic to
$\text{antisym*}(m,p^{2r})$. So $|N|=|\text{antisym*}(m,p^{2r})|=p^{rm^{2}}.$
Let

$O=\\{\begin{pmatrix}1&\mathbf{0}&\mathbf{0}\\\ \mathbf{0}&A&0_{n}\\\
\mathbf{0}&0_{n}&(\overline{A^{-1}})^{T}\end{pmatrix}:A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}})\\}.$

This is a subgroup of $U(2n+1,p^{2r})$ isomorphic to
$\text{Up}_{m}(\mathbb{F}_{p^{2r}})$. So
$|O|=|\text{Up}_{m}(\mathbb{F}_{p^{2r}})|=p^{rm(m-1)}.$ A direct computation
shows that $P\cdot N\cdot O=(P\times N)\rtimes O$ and

$|(P\times N)\rtimes O|=p^{rn(n-1)/2}=|\text{Syl}_{p}(U(n,p^{2r}))|$

Thus

$\displaystyle(P\times N)\rtimes O$
$\displaystyle=\\{\begin{pmatrix}1&\mathbf{0}&\mathbf{x}\\\
\overline{\mathbf{x}}^{T}&\text{Id}_{n}&0_{n}\\\
\mathbf{0}&0_{n}&\text{Id}_{n}\end{pmatrix}\begin{pmatrix}1&\mathbf{0}&\mathbf{0}\\\
\mathbf{0}&A&0_{n}\\\
\mathbf{0}&0_{n}&(\overline{A^{-1}})^{T}\end{pmatrix}\begin{pmatrix}1&\mathbf{0}&\mathbf{0}\\\
\mathbf{0}&\text{Id}_{n}&B\\\
\mathbf{0}&0_{n}&\text{Id}_{n}\end{pmatrix}:\mathbf{x}\in\mathbb{F}_{p^{2r}}^{m},A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}}),B\in\text{antisym*}(m,p^{2r})\\}.$

is a $p$-Sylow of $U(2m+1,p^{2r})$.

∎

######  Corollary 7.16.

For $n=2m+1$, any $p$,

$\text{Syl}_{p}(U(n,p^{2r}))\cong\left((\mathbb{F}_{p^{2r}}^{+})^{m}\times\text{antisym*}(m,p^{2r})\right)\rtimes\text{Up}_{m}(\mathbb{F}_{p^{2r}})),$

where the action of $\text{Up}_{n}(\mathbb{F}_{p^{2r}})$ on
$\text{antisym*}(n,p^{2r})$ is given by

$A(B)=AB\overline{A}^{T}.$

and the action of $\text{Up}_{m}(\mathbb{F}_{p^{2r}})$ on
$(\mathbb{F}_{p^{2r}}^{+})^{m}$ is given by

$A(\mathbf{x})=\mathbf{x}\overline{A}^{T}.$

###  7.3 Calculating the center

####  7.3.1 $n=2m$

######  Proposition 7.17.

For $n=2m$, any $p$,

$\displaystyle Z(\text{Syl}_{p}(U(n,p^{2r})))$
$\displaystyle=\\{\begin{pmatrix}\text{Id}_{m}&D\\\
0_{n}&\text{Id}_{m}\end{pmatrix}:D=\begin{pmatrix}y&\mathbf{0}\\\
\mathbf{0}&0_{n-1}\end{pmatrix},y=-\overline{y})\\}\cong
G_{p^{r}}\cong(\mathbb{Z}/p\mathbb{Z})^{r}$

For the proof of Proposition 7.17, we need the following lemma:

######  Lemma 7.18.

Given $D\in\text{antisym*}(m,p^{2r})$,

$AD=D(\overline{A^{-1}})^{T}\text{ }\forall
A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}})\Leftrightarrow
D=\begin{pmatrix}y&\mathbf{0}\\\ \mathbf{0}&0_{m-1}\end{pmatrix}$

Granting this lemma, we can prove Proposition 7.17:

###### Proof of Proposition 7.17.

111When $p=2$, the negatives will go away, but the arguments will still hold.

$\displaystyle\text{Syl}_{p}(U(2m,p^{2r}))$
$\displaystyle=\\{\begin{pmatrix}A&0_{n}\\\
0_{n}&(\overline{A^{-1}})^{T}\end{pmatrix}\begin{pmatrix}\text{Id}_{n}&B\\\
0_{n}&\text{Id}_{n}\end{pmatrix}:A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}}),B\in\text{antisym*}(m,p^{2r})\\}$
$\displaystyle=\\{\begin{pmatrix}A&AB\\\
0_{n}&(\overline{A^{-1}})^{T}\end{pmatrix}:A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}}),B\in\text{antisym*}(m,p^{2r})\\}.$

Note that

$\displaystyle\begin{pmatrix}A&AB\\\
0_{n}&(\overline{A^{-1}})^{T}\end{pmatrix}^{-1}\begin{pmatrix}C&CD\\\
0_{n}&(\overline{C^{-1}})^{T}\end{pmatrix}\begin{pmatrix}A&AB\\\
0_{n}&(\overline{A^{-1}})^{T}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}A^{-1}CA&A^{-1}CAB+A^{-1}CD(\overline{A}^{-1})^{T}-B((\overline{A^{-1}CA})^{-1})^{T}\\\
0_{n}&((\overline{A^{-1}CA})^{-1})^{T}\end{pmatrix}$

So $\begin{pmatrix}C&CD\\\ 0_{n}&(C^{-1})^{T}\end{pmatrix}\in
Z(\text{Syl}_{p}(U(2m,p^{2r})))$ if and only if

$C\in Z(\text{Up}_{m}(\mathbb{F}_{p^{2r}}))=\\{\begin{pmatrix}1&0&x\\\
\mathbf{0}&\text{Id}_{n-2}&\mathbf{0}\\\ 0&\mathbf{0}&1\end{pmatrix}\\}$

and

$CD=CB+CA^{-1}D(\overline{A^{-1}})^{T}-B(\overline{C^{-1}})^{T}$

for all
$A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}}),B\in\text{antisym*}(m,p^{2r})$.
Choosing $A=\text{Id}_{n}$, we need $CD=CB+CD-B(C^{-1})^{T}$. So we must have
$CB=B(\overline{C^{-1}})^{T}$ for all $B\in\text{antisym*}(m,p^{2r})$. Write

$C=\begin{pmatrix}1&\mathbf{0}&x\\\ \mathbf{0}&\text{Id}_{n-2}&\mathbf{0}\\\
0&\mathbf{0}&1\end{pmatrix}\in
Z(\text{Up}_{m}(\mathbb{F}_{p^{2r}}))=Z(\text{Syl}_{p}(PSL(m,p^{2r}))).$

(We calculated this center in the section on $PSL(n,p^{r})$.) Then

$(\overline{C^{-1}})^{T}=\begin{pmatrix}1&\mathbf{0}&-\overline{x}\\\
\mathbf{0}&\text{Id}_{n-2}&\mathbf{0}\\\
0&\mathbf{0}&1\end{pmatrix}^{T}=\begin{pmatrix}1&\mathbf{0}&0\\\
\mathbf{0}&\text{Id}_{n-2}&\mathbf{0}\\\
-\overline{x}&\mathbf{0}&1\end{pmatrix}.$

Then for $B=(b_{i,j})\in\text{antisym*}(m,p^{2r})$,

$CB=\begin{pmatrix}b_{1,1}-x\overline{b_{1,n}}&b_{1,2}-x\overline{b_{2,n}}&&\cdots&b_{1,n-1}-x\overline{b_{n-1,n}}&b_{1,n}+xb_{n,n}\\\
-\overline{b_{1,2}}&b_{2,2}&b_{2,3}&&\cdots&b_{2,n}\\\
\vdots&&\ddots&&&\vdots\\\ -\overline{b_{1,n-1}}&&&\cdots&&b_{n-1,n}\\\
-\overline{b_{1,n}}&&&\cdots&-\overline{b_{n-1,n}}&b_{n,n}\end{pmatrix}$

and

$B(\overline{C^{-1}})^{T}=\begin{pmatrix}b_{1,1}-\overline{x}b_{1,n}&b_{1,2}&&\cdots&b_{1,n}\\\
-\overline{b_{1,2}}-\overline{x}b_{2,n}&b_{2,2}&b_{2,3}&\cdots&b_{2,n}\\\
\vdots&&\ddots&&\vdots\\\
-\overline{b_{1,n-1}}-\overline{x}b_{n-1,n}&-\overline{b_{2,n-1}}&&\cdots&b_{n-1,n}\\\
-\overline{b_{1,n}}-\overline{x}b_{n,n}&-\overline{b_{2,n}}&\cdots&-\overline{b_{n-1,n}}&b_{n,n}\end{pmatrix}$

######  Remark 38.

I do not need $m>2$ here like I did for the orthogonal groups.

So we must have $x=0$, and hence $C=\text{Id}_{n}$. So the other requirement
above becomes

$D=A^{-1}D(\overline{A^{-1}})^{T}\Leftrightarrow AD=D(\overline{A^{-1}})^{T}$

for all $A\in\text{Up}_{m}(\mathbb{F}_{p^{2r}})$. Then by Lemma 7.18, we get
that

$Z(\text{Syl}_{p}(U(2m,p^{2r})))=\\{\begin{pmatrix}\text{Id}_{m}&D\\\
0_{n}&\text{Id}_{m}\end{pmatrix}:D=\begin{pmatrix}y&\mathbf{0}\\\
\mathbf{0}&0_{n-1}\end{pmatrix}\\}$

∎

###### Proof of Lemma 7.18.

$\Leftarrow$: This is a straightforward calculation.

$\Rightarrow$: We will prove this by induction.

Base Case: When $n=2$, we can write $A=\begin{pmatrix}1&a\\\ 0&1\end{pmatrix}$
and $D=\begin{pmatrix}y&x\\\ -\overline{x}&z\end{pmatrix}$.

$AD=\begin{pmatrix}y-a\overline{x}&x+az\\\ -\overline{x}&z\end{pmatrix},$

and

$D(\overline{A^{-1}})^{T}=\begin{pmatrix}y-\overline{a}x&x\\\
-\overline{x}-\overline{a}z&z\end{pmatrix}.$

So in order for these to be equal for all $A$, we must have $z=0$ and $x=0$.

Induction Step: The induction step is basically identical to that in the proof
of Lemma 5.8. ∎

####  7.3.2 $n=2m+1$

######  Proposition 7.19.

For $n=2m+1$, any $p$,

$\displaystyle Z(\text{Syl}_{p}(U(n,p^{2r})))$
$\displaystyle=\\{\begin{pmatrix}1&\mathbf{0}&\mathbf{x}\\\
\mathbf{x}^{T}&\text{Id}_{n}&D\\\
\mathbf{0}&0_{n}&\text{Id}_{n}\end{pmatrix}:\mathbf{x}=(x_{1},0,\cdots,0):D=\begin{pmatrix}y&\mathbf{0}\\\
\mathbf{0}&0_{n-1}\end{pmatrix},y=-\overline{y}))\\}$
$\displaystyle\cong\mathbb{F}_{p^{2r}}\times
G_{p^{r}}\cong(\mathbb{Z}/p\mathbb{Z})^{3r}$

###### Proof.

Since
$\text{Syl}_{p}(U(2m+1,p^{2r}))\cong\left((\mathbb{F}_{p^{2r}}^{+})^{m}\times(\text{antisym*}(m,p^{2r})\right)\rtimes\text{Up}_{m}(\mathbb{F}_{p^{2r}}))\cong(\mathbb{F}_{p^{2r}}^{+})^{n}\rtimes\text{Syl}_{p}(U(2m,p^{2r}))$,

${Z(\text{Syl}_{p}(U(2m+1,p^{2r})))}\cap\\{\mathbf{0}\\}\times{\text{Syl}_{p}(U(2m,p^{2r}))}$

must be a subset of $Z(\text{Syl}_{p}(U(2m,p^{2r}))$, which we proved above to
be

$Z(\text{Syl}_{p}(U(2m,p^{2r})))=\\{\begin{pmatrix}\text{Id}_{m}&D\\\
0_{m}&\text{Id}_{m}\end{pmatrix}:D=\begin{pmatrix}y&\mathbf{0}\\\
\mathbf{0}&0_{m-1}\end{pmatrix},y=-\overline{y}\\}$

Thus the center of $Z(\text{Syl}_{p}(U(2m+1,p^{r})))$ is a subset of

$(\mathbb{F}_{p^{2r}}^{+})^{m}\rtimes\left\\{\begin{pmatrix}\text{Id}_{m}&D\\\
0_{m}&\text{Id}_{m}\end{pmatrix}:D=\begin{pmatrix}y&\mathbf{0}\\\
\mathbf{0}&0_{m-1}\end{pmatrix}\right\\}=\left\\{\begin{pmatrix}1&\mathbf{0}&\mathbf{x}\\\
\mathbf{x}^{T}&\text{Id}_{m}&D\\\
\mathbf{0}&0_{m}&\text{Id}_{m}\end{pmatrix}:\mathbf{x}\in\mathbb{F}_{p^{2r}}^{m},D=\begin{pmatrix}y&\mathbf{0}\\\
\mathbf{0}&0_{m-1}\end{pmatrix}\right\\}$

Given $\begin{pmatrix}1&\mathbf{0}&\mathbf{y}\\\ \mathbf{y}^{T}&A&AB\\\
\mathbf{0}&0_{n}&(\overline{A^{-1}})^{T}\end{pmatrix}\in\text{Syl}_{p}(U(2m+1,p^{2r}))$,
we have that

$\begin{pmatrix}1&\mathbf{0}&\mathbf{x}\\\ \mathbf{x}^{T}&\text{Id}_{m}&D\\\
\mathbf{0}&0_{m}&\text{Id}_{m}\end{pmatrix}\begin{pmatrix}1&\mathbf{0}&\mathbf{y}\\\
\mathbf{y}^{T}&A&AB\\\
\mathbf{0}&0_{m}&(\overline{A^{-1}})^{T}\end{pmatrix}=\begin{pmatrix}1&\mathbf{0}&\mathbf{y}+\mathbf{x}(\overline{A^{-1}})^{T}\\\
\mathbf{x}^{T}+\mathbf{y}^{T}&A&\mathbf{x}^{T}y+AB+D(\overline{A^{-1}})^{T}\\\
\mathbf{0}&0_{m}&(\overline{A^{-1}})^{T}\end{pmatrix}$

and

$\begin{pmatrix}1&\mathbf{0}&\mathbf{y}\\\ \mathbf{y}^{T}&A&AB\\\
\mathbf{0}&0_{m}&(\overline{A^{-1}})^{T}\end{pmatrix}\begin{pmatrix}1&\mathbf{0}&\mathbf{x}\\\
\mathbf{x}^{T}&\text{Id}_{m}&D\\\
\mathbf{0}&0_{m}&\text{Id}_{m}\end{pmatrix}=\begin{pmatrix}1&\mathbf{0}&\mathbf{x}+\mathbf{y}\\\
\mathbf{y}^{T}+A\mathbf{x}^{T}&A&\mathbf{x}y^{T}+AD+AB\\\
\mathbf{0}&0_{m}&(\overline{A^{-1}})^{T}\end{pmatrix}$

So in order for

$\begin{pmatrix}1&\mathbf{0}&\mathbf{x}\\\ \mathbf{x}^{T}&\text{Id}_{m}&D\\\
\mathbf{0}&0_{n}&\text{Id}_{m}\end{pmatrix}$

to be in the center, we need $\mathbf{x}^{T}=A\mathbf{x}^{T}$,
$\mathbf{x}=\mathbf{x}(\overline{A^{-1}})^{T}$, and
$AD=D(\overline{A^{-1}})^{T}$ for all choices of $A$. By Lemma 7.18,
$AD=D(\overline{A^{-1}})^{T}$ is satisfied if and only if

$D=\begin{pmatrix}y&\mathbf{0}\\\ \mathbf{0}&0_{n-1}\end{pmatrix}.$

Note that the $k$th entry of $A\mathbf{x}^{T}$ is given by
$x_{k}+\sum_{i=k+1}^{n}x_{i}a_{k,i}$. In order for this to be equal to $x_{k}$
(the $k$th entry of $\mathbf{x}^{T}$) for all $a_{k,i}$, must have $x_{i}=0$
for all $i>1$. So $\mathbf{x}=(x_{1},0,\cdots,0)$. In this case
$\mathbf{x}=\mathbf{x}(\overline{A^{-1}})^{T}$ will be satisfied as well.
Therefore the center is

$\displaystyle Z(\text{Syl}_{p}(U(2m+1,p^{2r})))$
$\displaystyle=\\{\begin{pmatrix}1&\mathbf{0}&\mathbf{x}\\\
\mathbf{x}^{T}&\text{Id}_{m}&D\\\
\mathbf{0}&0_{m}&\text{Id}_{m}\end{pmatrix}:\mathbf{x}=(x_{1},0,\cdots,0):D=\begin{pmatrix}y&\mathbf{0}\\\
\mathbf{0}&0_{m-1}\end{pmatrix}\\}$

∎

###  7.4 Maximizing $L_{s}$

####  7.4.1 $n=2m$

Fix $\psi$ a non-trivial character of $\mathbb{F}_{p^{r}}^{+}$. By Lemma 3.8,
every element of

$\widehat{\text{antisym*}(m,p^{2r})}\cong\widehat{(\mathbb{F}_{p^{2r}}^{+})^{m(m-1)/2}\times\mathbb{F}_{p^{r}}^{m}}\cong\widehat{(\mathbb{F}_{p^{r}}^{+})^{m^{2}}}$

is of the form

$\mathbf{d}\mapsto\psi(\mathbf{b}\mathbf{d}^{T}),$

where $\mathbf{d},\mathbf{b}\in(\mathbb{F}_{p^{r}}^{+})^{m^{2}}$. Let the
isomorphism $\mathbb{F}_{p^{2r}}^{+}\to(\mathbb{F}_{p^{r}}^{+})^{2}$ be
defined as

$\dot{b}+\ddot{b}\alpha\mapsto(\dot{b},\ddot{b}).$

Note that

$\overline{(\dot{b},\ddot{b})}=\begin{cases}(\dot{b},-\ddot{b})&p\neq 2\\\
(\dot{b}+\ddot{b},\ddot{b})&p=2\end{cases}.$

And for $b=(\dot{b},\ddot{b})$,

$b+\overline{b}=\begin{cases}(2\dot{b},0)&p\neq 2\\\
(\dot{b},0)&p=2\end{cases}.$

Note that for a matrix $(b_{i,j})$ in $\text{Antisym}(m,p^{2r})$, $b_{i,i}\in
G_{p^{r}}$, so

$b_{i,i}=\begin{cases}\ddot{b_{i,i}}\alpha&p\neq 2\\\
\dot{b_{i,i}}&p=2\end{cases}.$

Let the isomorphism
$\text{antisym*}(m,p^{2r})\cong(\mathbb{F}_{p^{r}}^{+})^{m^{2}}$ be given by

$\displaystyle\begin{pmatrix}b_{1,1}&b_{1,2}&b_{1,3}&\cdots&b_{1,n}\\\
-\overline{b_{1,2}}&b_{2,2}&b_{2,3}&\cdots&b_{2,n}\\\
\vdots&&\ddots&&\vdots\\\
-\overline{b_{1,n-1}}&-\overline{b_{2,n-1}}&\cdots&b_{n-1,n-1}&b_{n-1,n}\\\
-\overline{b_{1,n}}&-\overline{b_{2,n}}&\cdots&-\overline{b_{n-1,n}}&b_{n,n}\end{pmatrix}\mapsto\begin{cases}(\ddot{b_{1,1}},\dot{b_{1,2}},\ddot{b_{1,2}},\cdots,\dot{b_{1,n}},\ddot{b_{1,n}},\ddot{b_{2,2}},\dot{b_{2,3}},\ddot{b_{2,3}},\cdots,\cdots,\ddot{b_{n,n}})&p\neq
2\\\
(\dot{b_{1,1}},\dot{b_{1,2}},\ddot{b_{1,2}},\cdots,\dot{b_{1,n}},\ddot{b_{1,n}},\dot{b_{2,2}},\dot{b_{2,3}},\ddot{b_{2,3}},\cdots,\cdots,\dot{b_{n,n}})&p=2\end{cases}$

######  Proposition 7.20.

For any $p$, the characters extending a non-trivial central character are
$\psi_{(b_{j})}$ with $b_{1}\neq 0$.

###### Proof.

$\psi_{(b_{j})}$ restricted to the center of the $p$-Sylow is $\psi_{b_{1}}$.
This is non-trivial if and only if $b_{1}\neq 0$. ∎

We will use Wigner-Mackey Theory with
$G=\text{Syl}_{p}(U(2m,p^{2r}))=\text{antisym*}(m,p^{2r})\rtimes\text{Up}_{m}(\mathbb{F}_{p^{2r}})$
to see what is the minimum dimension of an irreducible representation of $G$
with non-trivial central character. So we have
$\Delta=\text{antisym*}(m,p^{2r})\rtimes\\{\text{Id}_{m}\\}$,
$L=\\{0_{m}\\}\rtimes\text{Up}_{m}(\mathbb{F}_{p^{2r}})$. Recall that
$L_{s}=\text{Stab}_{L}(\psi_{s})$.

######  Remark 39.

We will view $\psi_{(b_{j})}$ as a map on
$\Delta\cong\text{antisym*}(m,p^{2r})\cong(\mathbb{F}_{p^{r}}^{+})^{m^{2}}$.
So
$\psi_{(b_{j})}(D,\text{Id})=\psi_{(b_{j})}(D)=\psi(\mathbf{b}\cdot\mathbf{d})$,
where $\mathbf{b}=(b_{j})$ and $\mathbf{d}$ is the vector corresponding to the
matrix $D$.

The action of $g\in\text{Syl}_{p}(U(2m,p^{r}))$ on $\widehat{\Delta}$ is given
by

${}^{g}\psi(D,\text{Id}_{m})=\psi(g^{-1}(D,\text{Id}_{m})g).$

So for $h=(0_{n},H^{-1})\in
L=\\{0_{m}\\}\times\text{Up}_{m}(\mathbb{F}_{p^{2r}})$, the action on
$\psi_{(b_{j})}$ is given by

${}^{h}\psi_{(b_{j})}(D,\text{Id}_{m})=\psi_{(b_{j})}((0_{n},H)(D,\text{Id}_{m})(0_{n},H^{-1})).$

So $(0_{n},H^{-1})\in L_{(b_{j})}$ if and only if for all
$\mathbf{d}\in((\mathbb{Z}/p\mathbb{Z})^{r})^{m^{2}}=D\in\text{antisym*}(m,p^{2r})$,

$\psi_{(b_{j})}((0_{n},H)(D,\text{Id}_{n})(0_{n},H^{-1}))=\psi_{(b_{j})}(D,\text{Id}_{n})$

Let $\mathbf{hdh^{T}}$ be the vector corresponding to $HDH^{T}$. Then since

$\psi_{(b_{j})}((0_{m},H)(D,\text{Id}_{m})(0_{m},H^{-1}))=\psi(\mathbf{b}\cdot\mathbf{hdh^{T}}),$

and

$\psi_{(b_{j})}(D,\text{Id}_{m})=\psi(\mathbf{b}\cdot\mathbf{d}).$

we get that $(0_{m},H^{-1})\in L_{s}$ if and only if

$\psi(\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d}))=1$

for all $\mathbf{d}\in(\mathbb{F}_{p^{r}})^{m^{2}}$ corresponding to
$D\in\text{antisym*}(m,p^{2r}).$

Write

$H=\begin{pmatrix}1&h_{1,2}&h_{1,3}&\cdots&h_{1,m}\\\
0&1&h_{2,3}&\cdots&h_{2,m}\\\ &&\ddots&&\vdots\\\ 0&0&\cdots&1&h_{m-1,m}\\\
0&0&0&\cdots&1\end{pmatrix},\qquad
D=\begin{pmatrix}d_{1,1}&d_{1,2}&d_{1,3}&\cdots&d_{1,m}\\\
-\overline{d_{1,2}}&d_{2,2}&d_{2,3}&\cdots&d_{2,m}\\\
\vdots&&\ddots&&\vdots\\\
-\overline{d_{1,n-1}}&-\overline{d_{2,m-1}}&\cdots&d_{m-1,m-1}&d_{m-1,m}\\\
-\overline{d_{1,m}}&-\overline{d_{2,m}}&\cdots&-\overline{d_{m-1,m}}&d_{m,m}\end{pmatrix}.$

######  Proposition 7.21.

For any $p$, $s=(b_{j})$, with $b_{1}\neq 0$,
$\max_{s}|L_{s}|=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}$, and
this maximum is achieved when $s=(b_{j})=(b_{1},0,\cdots,0)$ with $b_{1}\neq
0$.

######  Corollary 7.22.

For any $p$, for $s=(b_{1},0,0,\cdots)$ with $b_{1}\neq 0$, $\lambda$ trivial,

$\dim(\theta_{s,\lambda})=\frac{|L|}{|L_{s}|}=\frac{|\text{Up}_{m}(\mathbb{F}_{p^{2r}})|}{|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|}=|\mathbb{F}_{p^{2r}}^{(m-1)}|=p^{2r(m-1)}.$

###### Proof of Proposition 7.21.

Step 1: Proving that for any $\mathbf{s,|L_{s}|\leq p^{r(m-1)(m-2)}}$.

For $j_{0}>2$, choose $d_{i,j}=0$ except for
$d_{1,j_{0}}=-\overline{d_{j_{0},1}}$. Then we get that

$HD\overline{H}^{T}-D=\begin{pmatrix}\overline{h_{1,j_{0}}}d_{1,j_{0}}-h_{1,j_{0}}\overline{d_{1,j_{0}}}&\overline{h_{2,j_{0}}}d_{1,j_{0}}&\cdots&\overline{h_{j_{0}-1,j_{0}}}d_{1,j_{0}}&0&\cdots&0\\\
-h_{2,j_{0}}\overline{d_{1,j_{0}}}&0&&&\cdots&&0\\\
\vdots&&&\ddots&&&\vdots\\\
-h_{j_{0}-1,j_{0}}\overline{d_{1,j_{0}}}&0&&&\cdots&&0\\\ 0&&&&\cdots&&0\\\
\vdots&&&\ddots&&&\vdots\\\ 0&&&&\cdots&&0\end{pmatrix}$

Case 1: $\mathbf{p\neq 2}$

Note that

$\overline{(\dot{h_{1,j_{0}}}+\ddot{h_{1,j_{0}}}\alpha)}(\dot{d_{1,j_{0}}}+\ddot{d_{1,j_{0}}}\alpha)=(\dot{h_{1,j_{0}}}\dot{d_{1,j_{0}}}+\ddot{h_{1,j_{0}}}\ddot{d_{1,j_{0}}}\eta^{\prime})+(\dot{h_{1,j_{0}}}\ddot{d_{1,j_{0}}}-\ddot{h_{1,j_{0}}}\dot{d_{1,j_{0}}})\alpha$

So we have

$\mathbf{b}\cdot(\mathbf{hd\overline{h}^{T}}-\mathbf{d})=\dot{d_{1,j_{0}}}\left(2\dot{h_{1,j_{0}}}\ddot{b_{1,1}}+\sum_{i=2}^{j_{0}-1}(\dot{h_{i,j_{0}}}\dot{b_{1,i}}-\ddot{h_{i,j_{0}}}\ddot{b_{1,i}})\right)+\ddot{d_{1,j_{0}}}\left(2\ddot{h_{1,j_{0}}}\ddot{b_{1,1}}+\sum_{i=2}^{j_{0}-1}(\ddot{h_{i,j_{0}}}\eta^{\prime}\dot{b_{1,i}}+\dot{h_{i,j_{0}}}\ddot{b_{1,i}})\right)$

If
$2\dot{h_{1,j_{0}}}\ddot{b_{1,1}}+\sum_{i=2}^{j_{0}-1}(\dot{h_{i,j_{0}}}\dot{b_{1,i}}-\ddot{h_{i,j_{0}}}\ddot{b_{1,i}})\neq
0$, then as we run through all the values for $\dot{d_{1,j_{0}}}$ with
$\ddot{d_{1,j_{0}}}=0$, we will get that
$\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d})$ runs through all the values of
$\mathbb{F}_{p^{r}}$. And since $\psi$ is non-trivial, this means that
$\psi(\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d}))$ cannot always equal $1$.
This is a contradiction. So we must have

$2\dot{h_{1,j_{0}}}\ddot{b_{1,1}}+\sum_{i=2}^{j_{0}-1}(\dot{h_{i,j_{0}}}\dot{b_{1,i}}-\ddot{h_{i,j_{0}}}\ddot{b_{1,i}})=0$

for all choices of $j_{0}>1$. Similarly, by running through all values for
$\ddot{d_{1,j_{0}}}$ with $\dot{d_{1,j_{0}}}=0$, we see that we must have

$2\ddot{h_{1,j_{0}}}\ddot{b_{1,1}}+\sum_{i=2}^{j_{0}-1}(\ddot{h_{i,j_{0}}}\eta^{\prime}\dot{b_{1,i}}+\dot{h_{i,j_{0}}}\ddot{b_{1,i}})=0$

for all choices of $j_{0}>1$. If $b_{1,1}=\ddot{b_{1,1}}\alpha\neq 0$, then
given $h_{i,j_{0}}$ for $i>1$, $\dot{h_{1,j_{0}}}$ and $\ddot{h_{1,j_{0}}}$
are dictated by the equations above:

$\dot{h_{1,j_{0}}}=\frac{-1}{2\ddot{b_{1,1}}}\sum_{i=2}^{j_{0}-1}(\dot{h_{i,j_{0}}}\dot{b_{1,i}}-\ddot{h_{i,j_{0}}}\ddot{b_{1,i}})$

and

$\ddot{h_{1,j_{0}}}=\frac{-1}{2\ddot{b_{1,1}}}\sum_{i=2}^{j_{0}-1}(\ddot{h_{i,j_{0}}}\eta^{\prime}\dot{b_{1,i}}+\dot{h_{i,j_{0}}}\ddot{b_{1,i}}).$

Thus for $b_{1,1}\neq 0$, $p\neq 2$,

$|L_{s}|\leq|\\{H:H_{1,j}\text{ fixed }\forall j\neq
1\\}=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}$

This completes Step 1 for $p\neq 2$.

Case 2: $\mathbf{p=2}$

Note that

$\overline{(\dot{h_{1,j_{0}}}+\ddot{h_{1,j_{0}}}\alpha)}(\dot{d_{1,j_{0}}}+\ddot{d_{1,j_{0}}}\alpha)=(\dot{h_{1,j_{0}}}\dot{d_{1,j_{0}}}+\ddot{h_{1,j_{0}}}\dot{d_{1,j_{0}}}+\ddot{h_{1,j_{0}}}\ddot{d_{1,j_{0}}}\eta)+(\dot{h_{1,j_{0}}}\ddot{d_{1,j_{0}}}+\ddot{h_{1,j_{0}}}\dot{d_{1,j_{0}}})\alpha$

So we have

$\displaystyle\mathbf{b}\cdot(\mathbf{hd\overline{h}^{T}}-\mathbf{d})$
$\displaystyle=\dot{d_{1,j_{0}}}\left(\ddot{h_{1,j_{0}}}\ddot{b_{1,1}}+\sum_{i=2}^{j_{0}-1}(\dot{h_{i,j_{0}}}\dot{b_{1,i}}+\ddot{h_{1,j_{0}}}\dot{b_{1,i}}+\ddot{h_{i,j_{0}}}\ddot{b_{1,i}})\right)+\ddot{d_{1,j_{0}}}\left(\dot{h_{1,j_{0}}}\ddot{b_{1,1}}+\sum_{i=2}^{j_{0}-1}(\ddot{h_{i,j_{0}}}\eta\dot{b_{1,i}}+\dot{h_{i,j_{0}}}\ddot{b_{1,i}})\right)$

By the same reasoning as before, we must have

$\ddot{h_{1,j_{0}}}\ddot{b_{1,1}}+\sum_{i=2}^{j_{0}-1}(\dot{h_{i,j_{0}}}\dot{b_{1,i}}+\ddot{h_{1,j_{0}}}\dot{b_{1,i}}+\ddot{h_{i,j_{0}}}\ddot{b_{1,i}})=0$

and

$\dot{h_{1,j_{0}}}\ddot{b_{1,1}}+\sum_{i=2}^{j_{0}-1}(\ddot{h_{i,j_{0}}}\eta\dot{b_{1,i}}+\dot{h_{i,j_{0}}}\ddot{b_{1,i}})=0$

for all choices of $j_{0}>1$. If $b_{1,1}=\ddot{b_{1,1}}\alpha\neq 0$, then
given $h_{i,j_{0}}$ for $i>1$, $h_{1,j_{0}}^{1}$ and $h_{1,j_{0}}^{2}$ are
dictated by the equations above:

$\dot{h_{1,j_{0}}}=\frac{1}{\ddot{b_{1,1}}}\sum_{i=2}^{j_{0}-1}(\dot{h_{i,j_{0}}}\dot{b_{1,i}}+\ddot{h_{1,j_{0}}}\dot{b_{1,i}}+\ddot{h_{i,j_{0}}}\ddot{b_{1,i}})$

and

$\ddot{h_{1,j_{0}}}=\frac{1}{\ddot{b_{1,1}}}\sum_{i=2}^{j_{0}-1}(\ddot{h_{i,j_{0}}}\eta\dot{b_{1,i}}+\dot{h_{i,j_{0}}}\ddot{b_{1,i}}).$

Thus for $b_{1,1}\neq 0$,

$|L_{s}|\leq|\\{H:H_{1,j}\text{ fixed }\forall j\neq
1\\}=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}$

This completes Step 1 for $p=2$.

Step 2: Exhibiting that the max is achieved when
$\mathbf{s=(b_{1},0,\cdot,0)}$ with  $\mathbf{b_{1}\neq 0}.$

Let $s=(b_{1},0,0,\cdots,0)$. And let $B$ be the corresponding matrix. By the
calculation above, we must have that for $j_{0}>1$:

For $p\neq 2$,

$\dot{h_{1,j_{0}}}=\frac{1}{2\ddot{b_{1,1}}}\sum_{i=2}^{j_{0}-1}(\dot{h_{i,j_{0}}}\dot{b_{1,i}}-\ddot{h_{i,j_{0}}}\ddot{b_{1,i}})=0$

and

$\ddot{h_{1,j_{0}}}=\frac{1}{2\ddot{b_{1,1}}}\sum_{i=2}^{j_{0}-1}(\ddot{h_{i,j_{0}}}\eta^{\prime}\dot{b_{1,i}}+\dot{h_{i,j_{0}}}\ddot{b_{1,i}})=0.$

And for $p=2$,

$\dot{h_{1,j_{0}}}=\frac{1}{\ddot{b_{1,1}}}\sum_{i=2}^{j_{0}-1}(\dot{h_{i,j_{0}}}\dot{b_{1,i}}+\ddot{h_{1,j_{0}}}\dot{b_{1,i}}+\ddot{h_{i,j_{0}}}\ddot{b_{1,i}})=0$

and

$\ddot{h_{1,j_{0}}}=\frac{1}{\ddot{b_{1,1}}}\sum_{i=2}^{j_{0}-1}(\ddot{h_{i,j_{0}}}\eta\dot{b_{1,i}}+\dot{h_{i,j_{0}}}\ddot{b_{1,i}})=0.$

So, for any $p$ we have that for $j_{0}>0$,
$\dot{h_{1,j_{0}}}=0=\ddot{h_{1,j_{0}}}$, and hence $h_{1,j_{0}}=0$. So

$\displaystyle\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d})$
$\displaystyle=b_{1}(\sum_{l=1}^{n}[\overline{h_{1,l}}(\sum_{k=1}^{l}d_{k,l}h_{1,k}-\sum_{k=l+1}^{n}\overline{d_{l,k}}h_{1,k})]-d_{1,1})$
$\displaystyle\text{ since }b_{i}=0,i>1$
$\displaystyle=b_{1}(d_{1,1}-d_{1,1})$ $\displaystyle\text{ since
}h_{1,j}=0,j>1$ $\displaystyle=0$

Therefore,

$L_{s}=\\{H:H_{1,j}=0\text{ }\forall j>1\\}.$

∎

####  7.4.2 $n=2m+1$

Fix $\psi$ a non-trivial character of $\mathbb{F}_{p^{r}}^{+}$. By Lemma 3.8,
every element of

$\displaystyle\widehat{(\mathbb{F}_{p^{2r}}^{+})^{m}\times\text{antisym*}(m,p^{2r})}\cong\widehat{(\mathbb{F}_{p^{2r}}^{+})^{m}\times(\mathbb{F}_{p^{2r}}^{+})^{m^{2}}}\cong\widehat{(\mathbb{F}_{p^{r}}^{+})^{2m+m^{2}}}$

is of the form

$\mathbf{d}\mapsto\psi(\mathbf{b}\mathbf{d}^{T}),$

where $\mathbf{d},\mathbf{b}\in(\mathbb{F}_{p^{r}}^{+})^{2m+m^{2}}$. Let the
isomorphism $\text{antisym*}(m,p^{2r})\cong(\mathbb{F}_{p^{r}}^{+})^{m^{2}}$
be as it was for $n=2m$:

$\displaystyle\begin{pmatrix}b_{1,1}&b_{1,2}&b_{1,3}&\cdots&b_{1,n}\\\
-\overline{b_{1,2}}&b_{2,2}&b_{2,3}&\cdots&b_{2,n}\\\
\vdots&&\ddots&&\vdots\\\
-\overline{b_{1,n-1}}&-\overline{b_{2,n-1}}&\cdots&b_{n-1,n-1}&b_{n-1,n}\\\
-\overline{b_{1,n}}&-\overline{b_{2,n}}&\cdots&-\overline{b_{n-1,n}}&b_{n,n}\end{pmatrix}\mapsto\begin{cases}(\ddot{b_{1,1}},\dot{b_{1,2}},\ddot{b_{1,2}},\cdots,\dot{b_{1,n}},\ddot{b_{1,n}},\ddot{b_{2,2}},\dot{b_{2,3}},\ddot{b_{2,3}},\cdots,\cdots,\ddot{b_{n,n}})&p\neq
2\\\
(\dot{b_{1,1}},\dot{b_{1,2}},\ddot{b_{1,2}},\cdots,\dot{b_{1,n}},\ddot{b_{1,n}},\dot{b_{2,2}},\dot{b_{2,3}},\ddot{b_{2,3}},\cdots,\cdots,\dot{b_{n,n}})&p=2\end{cases}$

######  Proposition 7.23.

For any $p$, the characters of $(\mathbb{F}_{p^{r}}^{+})^{2m+m^{2}}$ extending
a non-trivial central character of the $p$-Sylow are $\psi_{(a_{j},b_{j})}$
with $(\dot{a_{1}},\ddot{a_{1}},b_{1})\neq(0,0,0)$.

###### Proof.

$\psi_{(a_{j},b_{j})}$ restricted to the center is
$\psi_{\dot{a_{1}},\ddot{a_{1}},b_{1}}$. This is non-trivial if and only if
$(\dot{a_{1}},\ddot{a_{1}},b_{1})\neq(0,0,0)$. ∎

We will use Wigner-Mackey Theory with
$G=\text{Syl}_{p}(U(2m+1,p^{r}))=\left((\mathbb{F}_{p^{2r}}^{+})^{m}\times\text{antisym*}(m,p^{2r})\right)\rtimes\text{Up}_{m}(\mathbb{F}_{p^{2r}})$
to see what is the minimum dimension of an irreducible representation of $G$
with non-trivial central character. So we have
$\Delta=\left((\mathbb{F}_{p^{2r}}^{+})^{m}\times\text{antisym*}(m,p^{2r})\right)\rtimes\\{\text{Id}_{n}\\}$,
$L=(\\{\mathbf{0}\\}\times\\{0_{m}\\})\rtimes\text{Up}_{m}(\mathbb{F}_{p^{2r}})$.
Recall that $L_{s}=\text{Stab}_{L}(\psi_{s})$.

######  Remark 40.

We will view $\psi_{(a_{j},b_{j})}$ as a map on
$\Delta\cong((\mathbb{F}_{p^{2r}}^{+})^{m}\times\text{antisym*}(n,p^{r}))\cong(\mathbb{Z}/p\mathbb{Z})^{2rm+rm^{2}}$.
So
$\psi_{(a_{j},b_{j})}(\mathbf{x},D,\text{Id})=\psi_{(a_{j},b_{j})}(\mathbf{x},D)=\psi((\mathbf{a},\mathbf{b})\cdot(\mathbf{d},\mathbf{x}))$,
where $(\mathbf{a},\mathbf{b})=(a_{j},b_{j})$ and $(\mathbf{x},\mathbf{d})$ is
the vector corresponding to $(\mathbf{x},D)$.

The action of $g\in\text{Syl}_{p}(U(2m+1,p^{2r}))$ on $\widehat{\Delta}$ is
given by

${}^{g}\psi(\mathbf{x},D,\text{Id}_{m})=\psi(g^{-1}(\mathbf{x},D,\text{Id}_{m})g).$

So for $h=(\mathbf{0},0_{m},H^{-1})\in
L=(\\{\mathbf{0}\\}\times\\{0_{m}\\})\rtimes\text{Up}_{m}(\mathbb{F}_{p^{2r}})$,
the action on $\psi_{(a_{j},b_{j})}$ is given by

${}^{h}\psi_{(a_{j},b_{j})}(\mathbf{x},D,\text{Id}_{m})=\psi_{(a_{j},b_{j})}((\mathbf{0},0_{m},H)(\mathbf{x},D,\text{Id}_{m})(\mathbf{0},0_{m},H^{-1})).$

So $(\mathbf{0},0_{m},H^{-1})\in L_{s}$ if and only if for all
$(\mathbf{x},\mathbf{d})\in(\mathbb{F}_{p^{r}}^{+})^{2m+m^{2}}=(\mathbf{x},D)\in(\mathbb{F}_{p^{2r}}^{+})^{m}\times\text{antisym*}(m,p^{2r})$,

$\psi_{(a_{j},b_{j})}((\mathbf{0},0_{n},H)(\mathbf{x},D,\text{Id}_{m})(\mathbf{0},0_{n},H^{-1}))=\psi_{(a_{j},b_{j})}(\mathbf{x},D,\text{Id}_{m})$

Let $\mathbf{hdh^{T}}$ be the vector corresponding to $HDH^{T}$. Then since

$\psi_{(a_{j},b_{j})}((\mathbf{0},0_{n},H)(\mathbf{x},D,\text{Id}_{m})(\mathbf{0},0_{n},H^{-1}))=\psi((\mathbf{a},\mathbf{b})\cdot(\mathbf{x}H^{T},\mathbf{hdh^{T}}))$

and

$\psi_{(a_{j},b_{j})}(\mathbf{x},D,\text{Id}_{n})=\psi((\mathbf{a},\mathbf{b})\cdot(\mathbf{x},\mathbf{d}))$

we get that $(\mathbf{0},0_{n},H^{-1})\in L_{s}$ if and only if

$\psi(\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})+\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d}))=1$

for all $(\mathbf{x},\mathbf{d})\in(\mathbb{F}_{p^{r}}^{+})^{2m+m^{2}}$
corresponding to
$(\mathbf{x},D)\in(\mathbb{F}_{p^{2r}}^{+})^{m}\times\text{antisym*}(m,p^{2r}).$

######  Proposition 7.24.

For any $p$, $s=(a_{j},b_{j})$,
$\max_{s}|L_{s}|=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}.$ And
this max is achieved when $s=(0,\cdots,0,b_{1},0,\cdots,0)$ and when
$s=(a_{1},0,\cdots,0)$ with $\dot{a_{1}}\neq 0,\ddot{a_{1}}=0$ or
$\dot{a_{1}}=0,\ddot{a_{1}}\neq 0$.

######  Corollary 7.25.

For any $p$, $\lambda$ trivial, if $s=(0,\cdots,0,b_{1},0,\cdots,0)$ or
$s=(a_{1},0,\cdots,0)$ with $\dot{a_{1}}\neq 0,\ddot{a_{1}}=0$ or
$\dot{a_{1}}=0,\ddot{a_{1}}\neq 0$, then

$\dim(\theta_{s,\lambda})=\frac{|L|}{|L_{s}|}=\frac{|\text{Up}_{m}(\mathbb{F}_{p^{2r}})|}{|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|}=|\mathbb{F}_{p^{2r}}^{(m-1)}|=p^{2r(m-1)}.$

###### Proof of Proposition 7.24.

Step 1: Proving that for any $\mathbf{s,|L_{s}|\leq p^{r(m-1)(m-2)}}$.

Case 1: $\mathbf{b_{1}\neq 0}$

If we take $\mathbf{x}=\mathbf{0}$, then
$\psi(\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})+\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d}))=1$
reduces to the condition for $U(2m,p^{2r})$. So if $b_{1}\neq 0$ we must have
$L_{s}=L_{(a_{j},b_{j})}$ a subset of the $L_{(b_{j})}$ calculated in the
proof of Proposition 7.21. I.e.

$L_{s}\subset\\{H:H_{1,j}\text{ fixed }\forall j\neq
1\\}=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}.$

Thus $|L_{s}|\leq p^{r(m-1)(m-2)}$.

Case 2: $\mathbf{a_{1}\neq 0}$ and $\mathbf{p\neq 2}$

If we take $\mathbf{d}=\mathbf{0}$, then
$\psi(\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})+\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d}))=1$
reduces to $\psi(\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x}))=1$. Note that

$x_{j_{0}}h_{k,j_{0}}=\dot{x_{j_{0}}}\dot{h_{k,j_{0}}}-\ddot{x_{j_{0}}}\ddot{h_{k,j_{0}}}\eta^{\prime}+(\dot{x_{j_{0}}}\ddot{h_{k,j_{0}}}+\ddot{x_{j_{0}}}\dot{h_{k,j_{0}}})\alpha$

So we have

$\displaystyle\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x}))$
$\displaystyle=\sum_{k=1}^{j_{0}-1}\dot{a_{k}}\dot{x_{j_{0}}h_{k,j_{0}}}+\ddot{a_{k}}\ddot{x_{j_{0}}h_{k,j_{0}}}$
$\displaystyle=\sum_{k=1}^{j_{0}-1}\dot{a_{k}}(\dot{x_{j_{0}}}\dot{h_{k,j_{0}}}-\ddot{x_{j_{0}}}\ddot{h_{k,j_{0}}}\eta^{\prime})+\ddot{a_{k}}(\dot{x_{j_{0}}}\ddot{h_{k,j_{0}}}+\ddot{x_{j_{0}}}\dot{h_{k,j_{0}}})$
$\displaystyle=\dot{x_{j_{0}}}(\sum_{k=1}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}})+\ddot{x_{j_{0}}}(\sum_{k=1}^{j_{0}-1}\ddot{a_{k}}\dot{h_{k,j_{0}}}-\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta^{\prime})$

If
$\sum_{k=1}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\neq
0$, then as we run through all the values for $\dot{x_{j_{0}}}$ with
$\ddot{x_{j_{0}}}=0$, we will get that
$\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x}))$ runs through all the values of
$\mathbb{F}_{p^{r}}$. And since $\psi$ is non-trivial, this means that
$\psi(\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})))$ cannot always equal $1$.
This is a contradiction. So we must have

$0=\sum_{k=1}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}=\dot{a_{1}}\dot{h_{1,j_{0}}}+\ddot{a_{1}}\ddot{h_{1,j_{0}}}+\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}$

Similarly, we must have

$0=\sum_{k=1}^{j_{0}-1}\ddot{a_{k}}\dot{h_{k,j_{0}}}-\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta^{\prime}=-\dot{a_{1}}\ddot{h_{1,j_{0}}}\eta^{\prime}+\ddot{a_{1}}\dot{h_{1,j_{0}}}+\sum_{k=2}^{j_{0}-1}\ddot{a_{k}}\dot{h_{k,j_{0}}}-\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta^{\prime}$

Subcase 2a: $\mathbf{\dot{a_{1}}\neq 0,\ddot{a_{1}}=0}$ or
$\mathbf{\dot{a_{1}}=0,\ddot{a_{1}}\neq 0}$

If $\dot{a_{1}}\neq 0$, $\ddot{a_{1}}=0$, then given $h_{k,j_{0}}$ for $k>1$,
$\dot{h_{1,j_{0}}}$ and $\ddot{h_{1,j_{0}}}$ are dictated by the above
equations:

$\displaystyle\dot{h_{1,j_{0}}}=\frac{-1}{\dot{a_{1}}}\left(\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)$

and

$\ddot{h_{1,j_{0}}}=\frac{1}{\dot{a_{1}}\eta^{\prime}}\left(\sum_{k=2}^{j_{0}-1}\ddot{a_{k}}\dot{h_{k,j_{0}}}-\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta^{\prime}\right)$

Similarly, if $\ddot{a_{1}}\neq 0$, $\dot{a_{1}}=0$, then given $h_{k,j_{0}}$
for $k>1$, $\dot{h_{1,j_{0}}}$ and $\ddot{h_{1,j_{0}}}$ are still dictated by
the above equations:

$\ddot{h_{1,j_{0}}}=\frac{-1}{\ddot{a_{1}}}\left(\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)$

and

$\dot{h_{1,j_{0}}}=\frac{-1}{\ddot{a_{1}}}\left(\sum_{k=2}^{j_{0}-1}\ddot{a_{k}}\dot{h_{k,j_{0}}}-\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta^{\prime}\right)$

Thus we have that for $\dot{a_{1}}\neq 0,\ddot{a_{1}}=0$ or
$\dot{a_{1}}=0,\ddot{a_{1}}\neq 0$,

$|L_{s}|\leq|\\{H:H_{1,j}\text{ fixed},\text{ }\forall j\neq
1\\}|=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}.$

Subcase 2b: $\mathbf{\dot{a_{1}},\ddot{a_{1}}\neq 0}$

If $\dot{a_{1}},\ddot{a_{1}}\neq 0$, then given $h_{k,j_{0}}$ for $k>1$, and
$\ddot{h_{1,j_{0}}}$, $\dot{h_{1,j_{0}}}$ is dictated:

$\dot{h_{1,j_{0}}}=\frac{-1}{\dot{a_{1}}}\left(\ddot{a_{1}}\ddot{h_{1,j_{0}}}+\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)$

and we must also have

$\dot{h_{1,j_{0}}}=\frac{-1}{\ddot{a_{1}}}\left(-\dot{a_{1}}\ddot{h_{1,j_{0}}}\eta^{\prime}+\sum_{k=2}^{j_{0}-1}\ddot{a_{k}}\dot{h_{k,j_{0}}}-\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta^{\prime}\right)$

Subtracting the first equation from the second we also get

$\displaystyle 0$
$\displaystyle=\ddot{h_{1,j_{0}}}(\frac{\ddot{a_{1}}}{\dot{a_{1}}}+\frac{\dot{a_{1}}\eta^{\prime}}{\ddot{a_{1}}})+\frac{1}{\dot{a_{1}}}\left(\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)-\frac{1}{\ddot{a_{1}}}\left(\sum_{k=2}^{j_{0}-1}\ddot{a_{k}}\dot{h_{k,j_{0}}}-\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta^{\prime}\right)$

If
$\frac{\ddot{a_{1}}}{\dot{a_{1}}}+\frac{\dot{a_{1}}\eta^{\prime}}{\ddot{a_{1}}}\neq
0$, we can solve for for $\ddot{h_{1,j_{0}}}$. And hence we would get that
$\ddot{h_{1,j_{0}}}$ is dictated by the above equations as well. So if
$\frac{\ddot{a_{1}}}{\dot{a_{1}}}+\frac{\dot{a_{1}}\eta^{\prime}}{\ddot{a_{1}}}\neq
0$, we again get that

$|L_{s}|\leq|\\{H:H_{1,j}\text{ fixed},\text{ }\forall j\neq
1\\}|=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}.$

Note that

$\frac{\ddot{a_{1}}}{\dot{a_{1}}}+\frac{\dot{a_{1}}\eta^{\prime}}{\ddot{a_{1}}}=0\Leftrightarrow(\frac{\ddot{a_{1}}}{\dot{a_{1}}})^{2}+\eta^{\prime}=0$

This will have a solution if and only if $f(x)=x^{2}+\eta^{\prime}$ has a
solution in $\mathbb{F}_{p^{r}}$. But we chose $\eta^{\prime}$ such that this
was irreducible over $\mathbb{F}_{p^{r}}$ and thus does not have any root in
$\mathbb{F}_{p^{r}}$. Therefore we can conclude that for any
$\dot{a_{1}},\ddot{a_{2}}\neq 0$, we also have

$|L_{s}|\leq|\\{H:H_{1,j}\text{ fixed},\text{ }\forall j\neq
1\\}|=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}.$

Case 3: $\mathbf{a_{1}\neq 0}$ and $\mathbf{p=2}$

As in Case 2, if we take $\mathbf{d}=\mathbf{0}$, then
$\psi(\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})+\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d}))=1$
reduces to $\psi(\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x}))=1$. Note that

$x_{j_{0}}h_{k,j_{0}}=\dot{x_{j_{0}}}\dot{h_{k,j_{0}}}+\ddot{x_{j_{0}}}\ddot{h_{k,j_{0}}}\eta+(\dot{x_{j_{0}}}\ddot{h_{k,j_{0}}}+\ddot{x_{j_{0}}}\dot{h_{k,j_{0}}}+\ddot{x_{j_{0}}}\ddot{h_{k,j_{0}}})\alpha$

So we have

$\displaystyle\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x}))$
$\displaystyle=\sum_{k=1}^{j_{0}-1}\dot{a_{k}}\dot{x_{j_{0}}h_{k,j_{0}}}+\ddot{a_{k}}\ddot{x_{j_{0}}h_{k,j_{0}}}$
$\displaystyle=\sum_{k=1}^{j_{0}-1}\dot{a_{k}}(\dot{x_{j_{0}}}\dot{h_{k,j_{0}}}+\ddot{x_{j_{0}}}\ddot{h_{k,j_{0}}}\eta)+\ddot{a_{k}}(\dot{x_{j_{0}}}\ddot{h_{k,j_{0}}}+\ddot{x_{j_{0}}}\dot{h_{k,j_{0}}}+\ddot{x_{j_{0}}}\ddot{h_{k,j_{0}}})$
$\displaystyle=\dot{x_{j_{0}}}(\sum_{k=1}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}})+\ddot{x_{j_{0}}}(\sum_{k=1}^{j_{0}-1}\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta+\ddot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}})$

By the same reasoning as before, we must have

$0=\sum_{k=1}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}=\dot{a_{1}}\dot{h_{1,j_{0}}}+\ddot{a_{1}}\ddot{h_{1,j_{0}}}+\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}$

and

$0=\sum_{k=1}^{j_{0}-1}\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta+\ddot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}=\dot{a_{1}}\ddot{h_{1,j_{0}}}\eta+\ddot{a_{1}}\dot{h_{1,j_{0}}}+\ddot{a_{1}}\ddot{h_{1,j_{0}}}+\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta+\ddot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}$

Subcase 3a: $\mathbf{\dot{a_{1}}\neq 0,\ddot{a_{1}}=0}$ or
$\mathbf{\dot{a_{1}}=0,\ddot{a_{1}}\neq 0}$

If $\dot{a_{1}}\neq 0$, $\ddot{a_{1}}=0$, then given $h_{k,j_{0}}$ for $k>1$,
$\dot{h_{1,j_{0}}}$ and $\ddot{h_{1,j_{0}}}$ are dictated by the above
equations:

$\displaystyle\dot{h_{1,j_{0}}}=\frac{1}{\dot{a_{1}}}\left(\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)$

and

$\ddot{h_{1,j_{0}}}=\frac{1}{\dot{a_{1}}\eta}\left(\sum_{k=2}^{j_{0}-1}\ddot{a_{k}}\dot{h_{k,j_{0}}}-\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta\right)$

Similarly, if $\ddot{a_{1}}\neq 0$, $\dot{a_{1}}=0$, then given $h_{k,j_{0}}$
for $k>1$, $\dot{h_{1,j_{0}}}$ and $\ddot{h_{1,j_{0}}}$ are still dictated by
the above equations:

$\ddot{h_{1,j_{0}}}=\frac{1}{\ddot{a_{1}}}\left(\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)$

and

$\dot{h_{1,j_{0}}}=\frac{1}{\ddot{a_{1}}}\left(\ddot{a_{1}}\ddot{h_{1,j_{0}}}+\sum_{k=2}^{j_{0}-1}\ddot{a_{k}}\dot{h_{k,j_{0}}}-\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta\right)$

Thus we have that for $\dot{a_{1}}\neq 0,\ddot{a_{1}}=0$ or
$\dot{a_{1}}=0,\ddot{a_{1}}\neq 0$,

$|L_{s}|\leq|\\{H:H_{1,j}\text{ fixed},\text{ }\forall j\neq
1\\}|=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}.$

Subcase 3b: $\mathbf{\dot{a_{1}},\ddot{a_{1}}\neq 0}$

If $\dot{a_{1}},\ddot{a_{1}}\neq 0$, then given $h_{k,j_{0}}$ for $k>1$ and
$\ddot{h_{1,j_{0}}}$, $\dot{h_{1,j_{0}}}$ is dictated:

$\dot{h_{1,j_{0}}}=\frac{1}{\dot{a_{1}}}\left(\ddot{a_{1}}\ddot{h_{1,j_{0}}}+\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)$

and we must also have

$\dot{h_{1,j_{0}}}=\frac{1}{\ddot{a_{1}}}\left(\dot{a_{1}}\ddot{h_{1,j_{0}}}\eta+\ddot{a_{1}}\ddot{h_{1,j_{0}}}+\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta+\ddot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)$

Adding these two equations we also get

$\displaystyle 0$
$\displaystyle=\ddot{h_{1,j_{0}}}(\frac{\ddot{a_{1}}}{\dot{a_{1}}}+\frac{\dot{a_{1}}\eta}{\ddot{a_{1}}}+1)+\frac{1}{\dot{a_{1}}}\left(\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)+\frac{1}{\ddot{a_{1}}}\left(\sum_{k=2}^{j_{0}-1}\dot{a_{k}}\ddot{h_{k,j_{0}}}\eta+\ddot{a_{k}}\dot{h_{k,j_{0}}}+\ddot{a_{k}}\ddot{h_{k,j_{0}}}\right)$

If
$\frac{\ddot{a_{1}}}{\dot{a_{1}}}+\frac{\dot{a_{1}}\eta}{\ddot{a_{1}}}+1\neq
0$, we can solve for for $\ddot{h_{1,j_{0}}}$. And hence we would get that
$\ddot{h_{1,j_{0}}}$ is dictated by the above equations as well. So if
$\frac{\ddot{a_{1}}}{\dot{a_{1}}}+\frac{\dot{a_{1}}\eta}{\ddot{a_{1}}}+1\neq
0$, we again get that

$|L_{s}|\leq|\\{H:H_{1,j}\text{ fixed},\text{ }\forall j\neq
1\\}|=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}.$

Note that

$\frac{\ddot{a_{1}}}{\dot{a_{1}}}+\frac{\dot{a_{1}}\eta}{\ddot{a_{1}}}+1=0\Leftrightarrow\ddot{a_{1}}^{2}+\dot{a_{1}}\ddot{a_{1}}+\dot{a_{1}}^{2}\eta=0.$

Recall that for $p=2$, we have

######  Proposition 7.26 ([Pom12], Proposition 1).

In a finite field of order $2^{r}$, for $f(x)=ax^{2}+bx+c$, we have have the
following:

  1. (i)

$f$ has exactly one root $\Leftrightarrow$ $b=0$.

  2. (ii)

$f$ has exactly two roots $\Leftrightarrow$ $b\neq 0$ and
$\text{Tr}(\frac{ac}{b^{2}})=0$.

  3. (iii)

$f$ has no root $\Leftrightarrow$ $b\neq 0$ and
$\text{Tr}(\frac{ac}{b^{2}})=1,$

where $\text{Tr}(x)=x+x^{2}+\cdots+x^{2^{r}-1}$.

So $\ddot{a_{1}}^{2}+\dot{a_{1}}\ddot{a_{1}}+\dot{a_{1}}^{2}\eta=0$ has a root
if and only if
$0=\text{Tr}(\frac{\dot{a_{1}}^{2}\eta}{\dot{a_{1}}^{2}})=\text{Tr}(\eta)$.
Recall that $\eta$ was chosen such that $x^{2}+x+\eta$ was irreducible, and
thus $x^{2}+x+\eta$ has no root. Hence by Proposition 7.26, we must have that
$\text{Tr}(\eta)=1$. Thus
$\ddot{a_{1}}^{2}+\dot{a_{1}}\ddot{a_{1}}+\dot{a_{1}}^{2}\eta=0$ has no root.
Therefore we can conclude that $\dot{a_{1}},\ddot{a_{2}}\neq 0$, we also have

$|L_{s}|\leq|\\{H:H_{1,j}\text{ fixed},\text{ }\forall j\neq
1\\}|=|\text{Up}_{m-1}(\mathbb{F}_{p^{2r}})|=p^{r(m-1)(m-2)}.$

Step 2: Exhibiting that the max is achieved when
$s=(0,\cdots,0,b_{1},0,\cdots,0)$ and when $s=(a_{1},0,\cdots,0)$ with
$\dot{a_{1}}\neq 0,\ddot{a_{1}}=0$ or $\dot{a_{1}}=0,\ddot{a_{1}}\neq 0$

Case 1: $\mathbf{b_{1}\neq 0}$

We showed in Step 1 that we must have $L_{(a_{j},b_{j})}$ a subset of the
$L_{(b_{j})}$ calculated for $n=2m$. Thus by Proposition 7.21 if
$(b_{j})=(b_{1},0,\cdots,0)$, then we have
$L_{s}\subset\\{H\in\text{Up}_{m}(\mathbb{F}_{p^{2r}}):H_{1,j}=0\text{
}\forall j>1\\}$ and for $H$ of this form, we have that
$\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d})=0$. Thus for $H$ of this form,
if $\mathbf{a}=0$,

$\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})+\mathbf{b}\cdot(\mathbf{hdh^{T}}-\mathbf{d})=0.$

So for $s=(0,\cdots,0,b_{1},0,\cdots,0)$,

$L_{s}=\\{H\in\text{Up}_{m}(\mathbb{F}_{p^{2r}}):H_{1,j}=0\text{ }\forall
j>1\\}.$

Case 2: $\mathbf{\dot{a_{1}}\neq 0,\ddot{a_{1}}=0}$ or
$\mathbf{\dot{a_{1}}=0,\ddot{a_{1}}\neq 0}$

If $a_{i}=0$ for $i\neq 1$, then by Subcase 2a and Subcase 3a in Step 1, we
get that for $j_{0}>1$, $\dot{h_{1,j_{0}}}=0=\ddot{h_{1,j_{0}}}$, and hence
$h_{1,j_{0}}=0$. So

$\displaystyle\mathbf{a}\cdot(\mathbf{x}H^{T}-\mathbf{x})$
$\displaystyle=\sum_{k=1}^{m-1}a_{k}\cdot(\sum_{j=k+1}^{m}x_{j}h_{k,j})$
$\displaystyle=a_{1}\cdot(\sum_{j=2}^{m}x_{j}h_{1,j}),$ $\displaystyle\text{
since }a_{i}=0,i>1$ $\displaystyle=0$ $\displaystyle\text{ since
}h_{1,j}=0,j>1$

So we get that for any $p$, $s=(a_{1},0,\cdots,0)$ with $\dot{a_{1}}\neq
0,\ddot{a_{1}}=0$ or $\dot{a_{1}}=0,\ddot{a_{1}}\neq 0$,

$L_{s}=\\{H:H_{1,j}=0,\text{ }\forall j\neq 1\\}.$

∎

###  7.5 Proof

###### Proof of Theorem 7.1.

######  Case 1 ($n=2$).

$\text{Up}_{1}(\mathbb{F}_{p^{2r}})$ is trivial and so
$\text{Syl}_{p}(U(2,p^{2r}))\cong\text{antisym*}(1,p^{2r})\cong\mathbb{F}_{p^{r}}^{+}$.
So
$\text{ed}_{k}(\text{Syl}_{p}(U(2,p^{2r})))=\text{ed}_{k}(\mathbb{F}_{p^{r}}^{+})=r$.
This proves Theorem 7.1 for $n=2$.

######  Case 2 ($n=3$).

$\text{Up}_{1}(\mathbb{F}_{p^{2r}})$ is trivial and so
$\text{Syl}_{p}(U(3,p^{2r}))\mathbb{F}_{p^{2r}}^{+}\times\text{antisym*}(1,p^{2r})\cong\mathbb{F}_{p^{2r}}^{+}\times\mathbb{F}_{p^{r}}^{+}\cong(\mathbb{Z}/p\mathbb{Z})^{3r}$.
So
$\text{ed}_{k}(\text{Syl}_{p}(U(3,p^{2r})))=\text{ed}_{k}((\mathbb{Z}/p\mathbb{Z})^{3r})=3r$.
This proves Theorem 7.1 for $n=3$.

######  Case 3 ($n=2m,m>1$).

Step 1: Find the lower bound

For any $p$, the minimum dimension of an irreducible representation of
$\text{Syl}_{p}(U(2m,p^{r}))$ with non-trivial central character is given by
$\theta_{s,\lambda}$ with $\lambda$ trivial and $s=(b,0,\cdots,0)$. So the
minimum dimension is $p^{r(n-1)}$. Thus the minimum dimension of a faithful
representation is $rp^{2r(m-1)}$ since the center has rank $r$.

Step 2: Construct the upper bound

Let $\\{e_{i}\\}$ be a basis for
$Z(U(2m,p^{2r}))\cong(\mathbb{Z}/p\mathbb{Z})^{r}$ over
$\mathbb{Z}/p\mathbb{Z}$. Then by Lemma 7, $\\{\psi_{e_{i}}\\}$ form a basis
for the characters of $Z(U(2m,p^{2r}))$. Thus
$\\{\psi_{e_{i}}|_{Z(U(2m,p^{2r}))}\\}$ form a basis for
$\widehat{(Z(U(2m,p^{2r}))}$. Let

$s_{i}=(e_{i},0,\cdots,0).$

Then by Corollary 7.22, for $\lambda$ trivial,
$\dim(\theta_{s_{i},\lambda})=p^{2r(m-1)}.$ Let

$\rho=\oplus_{i}\theta_{s_{i},\lambda}.$

Then

$\dim(\rho)=\sum_{i=1}^{r}\dim(\theta_{s_{i},\lambda})=\sum_{i=1}^{r}p^{2r(m-1)}=rp^{2r(m-1)}.$

And by Lemma 3.3, since $\psi_{e_{i}}|_{Z(U(2m,p^{2r}))}$ form a basis for
$\widehat{(Z(U(2m,p^{2r}))}$, we have that $\oplus_{i}\theta_{s_{i},\lambda}$
is a faithful representation of $\text{Syl}_{p}(U(2m,p^{2r}))$. Thus $\rho$ is
the desired faithful representation of $\text{Syl}_{p}(U(2m,p^{2r}))$ of
dimension $rp^{2r(m-1)}$.

######  Case 4 ($n=2m+1,m>1$).

Step 1: Find the lower bound

By Proposition 7.24, the minimum dimension of an irreducible representation
with non-trivial central character is $p^{2r(m-1)}$. Thus the minimum
dimension of a faithful representation is $3rp^{2r(m-1)}$ since the center has
rank $3r$.

Step 2: Construct the upper bound

Let $\\{e_{i}\\}$ be a basis for $(\mathbb{Z}/p\mathbb{Z})^{r}$ over
$\mathbb{Z}/p\mathbb{Z}$. Let

$S=\\{(0,\cdots,0,e_{i},0,\cdots,0),(e_{i},0,\cdots,0),(e_{i}\alpha,0,\cdots,0)\\}.$

Then $\\{\psi_{s}\\}_{s\in S}$ form a basis for the characters of
$Z(U(2m+1,p^{2r}))$. Thus $\\{\psi_{s}|_{\Omega_{1}(Z(U(2m+1,p^{2r})}\\}$ form
a basis for $\widehat{\Omega_{1}}(Z(U(2m+1,p^{2r})))$. By Corollary 7.25, for
$s\in S$, $\lambda$ trivial,

$\dim(\theta_{s,\lambda})=p^{2r(m-1)}$

Let $\rho=\oplus_{s\in S}\theta_{s,\lambda}$. Then

$\dim(\rho)=\sum_{s}\dim(\theta_{s,\lambda})=3rp^{2r(m-1)}$

And by Lemma 3.3, since $\psi_{s}|_{\Omega_{1}(Z(U(2m+1,p^{2r})}$ form a basis
for $\widehat{\Omega_{1}}(Z(U(2m+1,p^{2r})))$, we have that $\oplus_{s\in
S}\theta_{s,\lambda}$ is a faithful representation of
$\text{Syl}_{p}(U(2m+1,p^{2r}))$. Thus $\rho$ is the desired faithful
representation of $\text{Syl}_{p}(U(2m+1,p^{2r}))$ of dimension
$3rp^{2r(m-1)}$.

∎

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