Consider the Lie group ${\rm Sp}(6,\mathbb{R})$ defined as

${\rm Sp}(6,\mathbb{R}) = \{g\in {\rm GL}(6,\mathbb{R}): {}^tgJg=J \},$

where

$J=\begin{bmatrix} &&&&&1\\ &&&&1&\\ &&&1&&\\ &&-1&&&\\ &-1&&&&\\ -1&&&&&\\ \end{bmatrix}.$

Its Lie algebra $sp(6,\mathbb{R})$ is the set of all $6\times 6$ matrices $X$ with real entries such that

${}^tXJ+JX=0.$

A general element of $sp(6,\mathbb{R})$ is of the form

$\begin{bmatrix} a & b & c & d & e & f\\ g & h & i & j & k & e\\ l & m & n & o & j & d\\ p & q & r & -n & -i & -c\\ s & t & q & -m & -h & -b\\ u & s & p & -l & -g & -a\\ \end{bmatrix}$

and so $sp(6,\mathbb{R})$ is a 21-dimensional real vector space. We would like to describe its root system.

A basis for this space is given below.

$H_1=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -1\\ \end{bmatrix} H_2=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$

$H_3=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix},$

$X_1=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, X_1'={}^tX_1,$

$X_2=\begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, X_2'={}^tX_2$

$X_3=\begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, X_3'={}^tX_3,$

$X_4=\begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, X_4'={}^tX_4$

$X_5=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, X_5'={}^tX_5,$

$X_6=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, X_6'={}^tX_6$

$X_7=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, X_7'={}^tX_7,$

$X_8=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, X_8'={}^tX_8$

$X_9=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, X_9'={}^tX_9$

We complexify the Lie algebra by tensoring with $\mathbb{C}$ to get $sp(6,\mathbb{C})=sp(6,\mathbb{R}\otimes\mathbb{C}$. A basis for the complexified Lie algebra is given below. We use the notation $\overline{X}=(\overline{c_{ij}})$ to indicate the matrix whose $ij$-th entry is the complex conjugate of the $ij$-th entry of $X$.

$Z_1=-i\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, Z_2=-i\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix},$

$Z_3=-i\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$

$Y_{1+}=\dfrac{1}{2}\begin{bmatrix} 0 & 1 & 0 & 0 & -i & 0\\ -1 & 0 & 0 & 0 & 0 & -i\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ i & 0 & 0 & 0 & 0 & -1\\ 0 & i & 0 & 0 & 1 & 0\\ \end{bmatrix}, Y_{1-}=\overline{Y_{1+}},$

$Y_{2+}=\dfrac{1}{2}\begin{bmatrix} 0 & 0 & 1 & -i & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0 & 0 & -i\\ i & 0 & 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & i & 1 & 0 & 0\\ \end{bmatrix}, Y_{2-}=\overline{Y_{2+}}$

$Y_{3+}=\dfrac{1}{2}\begin{bmatrix} 0 & 0 & 1 & i & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & i\\ i & 0 & 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & i & -1 & 0 & 0\\ \end{bmatrix}, Y_{3-}=\overline{Y_{3+}},$

$Y_{4+}=\dfrac{1}{2}\begin{bmatrix} 0 & 1 & 0 & 0 & i & 0\\ 1 & 0 & 0 & 0 & 0 & i\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ i & 0 & 0 & 0 & 0 & -1\\ 0 & i & 0 & 0 & -1 & 0\\ \end{bmatrix}, Y_{4-}=\overline{Y_{4+}}$

$Y_{5+}=\dfrac{1}{2}\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & -i & 0 & 0\\ 0 & -1 & 0 & 0 & -i & 0\\ 0 & i & 0 & 0 & -1 & 0\\ 0 & 0 & i & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, Y_{5-}=\overline{Y_{5+}},$

$Y_{6+}=\dfrac{1}{2}\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & i & 0 & 0\\ 0 & 1 & 0 & 0 & i & 0\\ 0 & i & 0 & 0 & -1 & 0\\ 0 & 0 & i & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, Y_{6-}=\overline{Y_{6+}}$

$Y_{7+}=\dfrac{1}{2}\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & i\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ i & 0 & 0 & 0 & 0 & -1\\ \end{bmatrix}, Y_{7-}=\overline{Y_{7+}},$

$Y_{8+}=\dfrac{1}{2}\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & i & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & i & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, Y_{8-}=\overline{Y_{8+}}$

$Y_{9+}=\dfrac{1}{2}\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & i & 0 & 0\\ 0 & 0 & i & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}, Y_{9-}=\overline{Y_{9+}}$

If we choose $Z_1, Z_2, Z_3$ as a Cartan subalgebra, then we get the following picture of the root system of $sp(6,\mathbb{C})$:

Definition: We say that a vector $v$ in a representation of $\mathfrak{g}_\mathbb{C}$ has weight $(l_1,l_2,l_3)\in \mathbb{Z}^3$ if

$Z_1v=l_1v, Z_2v=l_2v, Z_3v=l_3v$

The arrows in our picture of the root system indicate how the elements of $\mathfrak{g}_\mathbb{C}$ change the weights. For example, when $Y_{1+}$ acts on a vector with weight $(l_1,l_2,l_3)$, it changes the weight to $(l_1+1,l_2-1,l_3)$. $Y_{6-}$ changes the weight to $(l_1,l_2-1,l_3-1)$.

If the representation comes from a representation of $G$ with trivial central character, then the sum $l_1+l_2+l_3$ must be even. So either all the weight are even or there are precisely two that are odd. Indeed, we have

$exp(t\begin{bmatrix} &1\\ -1 & \\ \end{bmatrix})=\begin{bmatrix} \cos t & \sin t\\ -\sin t & \cos t \end{bmatrix}.$

In particular,
$exp(\pi\begin{bmatrix} &1\\ -1 & \\ \end{bmatrix})=\begin{bmatrix} -1 & \\ & -1 \end{bmatrix}.$

$(-I_6)v=\begin{bmatrix} \cos\pi & & & \sin\pi & &\\ & \cos\pi & & & \sin\pi & \\ & & \cos\pi & & & \sin\pi\\ -\sin\pi & & &\cos\pi & &\\ & -\sin\pi & & &\cos\pi &\\ & & -\sin\pi & & & \cos\pi\\ \end{bmatrix}v$

$=e^{il_1\pi}e^{il_2\pi}e^{il_3\pi}v=(-1)^{l_1+l_2+l_3}v=v$ (since the center acts trivially.)