# A basic example to understand the concept of “Weight”

Let $$A=b(2,\mathbb{R})$$ be he Lie subalgebra of upper triangle matrices of $$gl(2,\mathbb{R})$$. It is clear that $$e_1=(1,0)$$ is an eigenvector for $$A$$, because it is an eigenvector for every element of $$A$$; that is, $$a(v) \in Span\{(1,0)\}$$ for every $$a \in A$$. Now I appreciate your answer to find the corresponding weight to $$(1,0)$$ and determine its weight space. Recall that we have the definition of weight as

A weight for a Lie subalgebra $$A$$ of $$gl(V)$$ is a linear map $$\lambda: A \to F$$ such that $$V_{\lambda}= \{v \in V: av=\lambda (a) v ~~ \text{for all}~ a \in A \}$$ is a non-zero subspace pf $$V$$.

• Just a point to make this precise: I suppose you are looking at the defining ($2$-dimensional) representation of the Lie algebra $gl(2,\Bbb R)$ (which is then restricted to the subalgebra $A$). I might be interesting to begin writing down the definition of this representation. – Marc van Leeuwen Apr 18 at 17:24
• I am looking at the defining (2-dimensional) representation as well, but unfortunately I do not understand the weight concept precisely! and as a consequence the concept of highest weight vectors for more studies. I appreciate your answer for a small example to clarify this concepts for me. – user40491 Apr 18 at 17:29

You seem to be hesitant to write down the definition of the action, which is simply $$M.v=Mv$$ (matrix times vector multiplication) for $$v\in\Bbb R^2$$ and $$M\in gl(2,\Bbb R)$$. (In your definition of $$V_\lambda$$, the action $$M.v$$ is written as $$M(v)$$, which I find slightly less proper as elements of the Lie algebra $$gl(2,\Bbb R)$$ are not functions taking a vector of some representation as argument.) Now if $$M={a~~b\choose0~~c}$$ and $$v=e_1$$ you can check that $$M.v=av$$, so the eigenvalue of $$e_1$$ for $$M$$ is $$a$$. The weight $$\lambda$$ of $$e_1$$ is the map $$A\to\Bbb R$$ given by $$M\mapsto a$$ that takes the top-left entry of the matrix; this is indeed a linear map on $$A$$. If you like, $$\lambda$$ is defined by $$\lambda({a~~b\choose0~~c})=a$$.
For this $$\lambda$$ the weight space $$V_\lambda$$ is $$\{\,v\in V\mid\forall{a~~b\choose0~~c}\in gl(2,\Bbb R):{a~~b\choose0~~c}.v=av\,\}$$ which is the span $$\langle e_1\rangle$$ of $$e_1$$ (even for a single such matrix with $$b\neq0$$ or $$c\neq a$$, the eigenspace for $$a$$ is just $$\langle e_1\rangle$$ so $$V_\lambda$$ is certainly contained in this subspace; and every vector $$\langle e_1\rangle$$ satisfies the condition for being in $$V_\lambda$$). It happens that this is the only weight space for this representation: most matrices in $$A$$ are not diagonalisable and have just this one eigenspace. (It is more customary to consider weight spaces for Abelian subalgebras of semisimple elements, such as the diagonal matrices in $$gl(2,\Bbb R)$$; then the weight spaces span the whole representation space.)
• Let consider your definition of $\lambda$ then weight space is $\mathbb{R}$? – user40491 Apr 18 at 18:19
• @user40491 No you need a subspace of $V$. See my extended answer. – Marc van Leeuwen Apr 18 at 19:44