The only even weights occurring in $M$ are $0, \pm 2$? The source of this question is Humphreys' Introduction to Lie Algebras and Representation Theory.
I'm having trouble understanding the last paragraph on page 38 in section 8.4 Integrality properties. It goes like this:
$(1)$ $L$ is a finite-dimensional semisimple Lie algebra over $F$, $\alpha \in \Phi$ is a root, $S_\alpha =F\{ x_\alpha, y_\alpha, h_\alpha \}$ is a copy of $\mathfrak {sl}(2,F)$ inside $L$, and $M$ is the subspace of $L$ spanned by $H=L_0$ and all the root spaces $L_{c\alpha}$ for $0 \neq c \in F$. $M$ is an $S_\alpha$-submodule of $L$ and the weights of $h_\alpha$ on $M$ are integers $0$ and $2c=c\alpha (h_\alpha)$.
This is all fine with me. Then it says:
$(2)$ $S_\alpha$ acts trivially on $\ker(\alpha)$, which is a subspace of codimension $1$ in $H$ complementary to $Fh_\alpha$, while $S_\alpha$ is an irreducible $S_\alpha$-submodule of $M$.
I believe this, but I don't understand the purpose of saying it for what follows:
$(3)$ Taken together, $\ker(\alpha)$ and $S_\alpha$ exhaust all occurrences of the weight $0$ for $h_\alpha$. So the only even weights occurring in $M$ are $0, \pm 2$. 
This is where I'm having trouble. First of all, doesn't $H$ exhaust all occurrences of the weight $0$ for $h_\alpha$? Next, I don't understand why other even weights for $h_\alpha$ like $\pm 4,6,...$ can't appear in $M$. Lastly, what do the statements in $(2)$ have to do with the conclusion in $(3)$?
Thanks.
 A: To understand this part, you need to be familiar with representation of $\mathfrak{sl}(2)$.
 From section 7, we know that any finite dimensional representation of $\mathfrak{sl}(2)$ is completely reducible, and each irreducible component has either a weight (with respect to $h$, the semisimple element in $\mathfrak{sl}(2)$) $0$, or a weight $1$. And the number of irreducible summands with even weight is equal to $\dim(V_0)$, the number of irreducible summands with odd weight is equal to $\dim(V_1)$. Here, $V_0$ and $V_1$ is the notation used in section 7 of Humphreys. 
Now, in your case, $M=H\oplus (\oplus_{c\in\mathbb{C}^*} L_{c\alpha})$ is a $\mathfrak{sl}(2)$-module. 


*

*I claim that $V_0=H$(this is easy to check), thus the number of the irreducible summands with even weights is $\dim(V_0)=\dim(H)$. 

*Consider $\ker(\alpha)\subset H$, which is a subspace of $H$ with dimension $\dim(H)-1$. On the other hand, $\ker(\alpha)$ is also a trivial $S_{\alpha}$-module, since $x_{\alpha}.h=[x_{\alpha},h]=-[h,x_{\alpha}]=-\alpha(h)x_{\alpha}=0, \forall h\in \ker(\alpha)$.(Similarly, $y_{\alpha}$ also acts trivially on this module, $h_\alpha$ acts trivially because $H$ is abelian.) Therefore, $\ker(\alpha)$ is a direct sum of $(\dim(H)-1)$-trivial modules. 

*From 1 and 2 above, we know that the number of "irreducible summands with even weight" is $\dim(H)$ and we've found $(\dim(H)-1)$ of these summands are trivial. Now, notice that $S_{\alpha}\subset M$ is also an irreducible submodule with even weight (which are $2,0,-2$). So, if we put the non-trivial irreducible submodule $S_{\alpha}$ and $\ker(\alpha)$ together, there will be $1+(\dim(H)-1)=\dim(H)=\dim(V_0)$ irreducible summands with even weight, which exhaust all of the occurrences of weight zero for $h_\alpha$. Thus, until now, we've found all of the irreducible summands of $M$ with even weights. And among these weights, we've seen that only $2,0,-2$ occurs, there's no weight $4,6,8$, etc.
Answer to your first question: Yes, $H$ does exhaust all occurrences of weight $0$ for $h_\alpha$, but $H$ itself is not a submodule of $M$. The author means $S_{\alpha}$ together with $\ker(\alpha)$ exhaust the irreducible summands with even weights. 
Answer to your second question: This is explained in step 3 above. Since the submodule $S_{\alpha}\subset M$ only provides the weight $2,0,-2$, and $\ker(\alpha)$ only provides the weight $0$, so there's no other even weights. 
And I think now you know the answer to your third question.
