**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.




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**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 $W$.




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**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.



Show that if $\Theta$ is an infinitesimal weight of a real $T$-module $W$ ($T$ is a torus) then $-\Theta$ is also a weight
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 $L\_X \bar{w}\_j=L\_X u\_j - \sqrt{-1}L\_X v\_j=\overline{L\_X w\_j}=\overline{\Theta\_j(X)w\_j}=-\Theta\_j(X)\bar{w}\_j$.


