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.)

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