Show that if $\Theta$ is an infinitesimal weight of a real $T$-module $W$ ($T$ is a torus) then $-\Theta$ is also a weight.

It is an exercise of Bröcker's book on Representations of Compact Lie Groups.

Here is some language

Let $T$ be a torus and $LT$ be its Lie algebra. A weight of a complex $T$-module $V$ is a irreducible character $\chi:T\to U(1)$ s.t. the corresponding weight space $$V(\chi)=\{v\in V: x\cdot v=\chi(x)v, \,\forall x\in T\}$$ is nonzero.

Its differential $\Theta=d\chi:LT\to LU(1)\cong i\mathbb{R}$ is called a infinitesimal weight of $V$ if $$V(\Theta)=\{v\in V: L_X v=\Theta(X)v, \,\forall X\in LT\}$$ is nonzero.

$L_X$ stands for the lie derivative $L_Xv:=\lim_{t\to 0}t^{-1}(exp(tX)\cdot v-v)$.

We have that every complex $T$-module $V$ decomposes into a finite sum of weight spaces $$V=\oplus_j V(\Theta_j).$$

An infinitesimal weight of a real $T$-module $W$ is a infinitesimal weight of its complexification $W_\mathbb{C}:= \mathbb{C}\otimes_\mathbb{R} W$.

As $W_\mathbb{C}$ decomposes into a sum of weight spaces, so does $W$.

Here is what I'm trying to use

Let $\Theta_1$ be a infinitesimal weight of the real $T$-module $W=\oplus_j W(\Theta_j)$.

Take a $w\in W\setminus{0}$. Then $w=\sum_{j}w_j$, with $w_j\in W(\Theta_j)$. Let $k$ be a index s.t. $w_k\neq 0$.

Note that $$L_X w=\sum_j L_Xw_j=\sum_j\Theta_j(X)w_j.$$

So $w\in W(-\Theta_1)$ iff $$-\Theta_1(X)w=\sum_j-\Theta_1(X)w_j=L_Xw=\sum_j\Theta_j(X)w_j;$$ for every $X\in LT$.

As $w_k\neq 0$, $\Theta_k=-\Theta_1$.

But this approach doesn't seem very useful.

  • $\begingroup$ Have you tried taking complex conjugate of $w_k$? $\endgroup$
    – Ben McKay
    Commented Jan 29, 2019 at 10:54
  • $\begingroup$ @BenMcKay I can't see how this help, could you elucidate a little more? $\endgroup$
    – Gomes93
    Commented Jan 29, 2019 at 12:51
  • $\begingroup$ Do you mean the derivative of $\overline{\chi}$? $\endgroup$
    – Gomes93
    Commented Jan 29, 2019 at 15:01

1 Answer 1


Since $W$ is a real vector space, $L_X$ is a real operator, so if we write $w_j=u_j+\sqrt{-1}v_j$, then $L_X \bar{w}_j=L_X u_j - \sqrt{-1}L_X v_j=\overline{L_X w_j}=\overline{\Theta_j(X)w_j}=-\Theta_j(X)\bar{w}_j$.


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