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

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.

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

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 $$L_X \bar{w}_j=L_X u_j - \sqrt{-1}L_X v_j=\overline{L_X w_j}=\overline{\Theta_j(X)w_j}=-\Theta_j(X)\bar{w}_j$$.