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Here's a link to the PDF. I'm using the 2nd edition, typeset using Latex. In section 2.4, page 37, Samelson proves:

PROPOSITION A. For each $\alpha$ in $\Delta$ the subspace $[\mathfrak g_{\alpha}, \mathfrak g_{-\alpha}]$ of $\mathfrak h$ has dimension 1, and the restriction of $\alpha$ to it is not identically $0$.

$\Delta$ above is the set of roots of a semisimple Lie algebra $\mathfrak g$. The Cartan subalgebra is $\mathfrak h$. $\mathfrak g_{\alpha}$ is vector subspace $\{X \in \mathfrak g \mid (\forall H \in H) [H,X]=\alpha(H)X\}$. The problem is that Samelson only seems to prove that $\dim([\mathfrak g_{\alpha}, \mathfrak g_{-\alpha}]) > 0$, and not the stronger fact that $\dim[\mathfrak g_{\alpha}, \mathfrak g_{-\alpha}]=1$. I'm wondering how to prove this.

Additionally, I must remark that there appear to be typos where Samelson seems to define "$\alpha$-string of $\beta$" inconsistently: At some point, he talks about roots of the form $\beta + t\alpha$ (which I think he intends to mean throughout) but he also defines such a string in terms of $\alpha + t\beta$ (which I think is a typo). Additionally, I suspect that $Y$ at the bottom of the page should belong to $\mathfrak g_{-\alpha}$ instead of $\mathfrak g_{\alpha}$.

The confusion here is such that I feel stuck, and find it hard to carry on with much confidence through the rest of the book.

wlad
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    Agreed that $Y$ should be in $\mathfrak g_{-\alpha}$, and that $\alpha +t\beta$ is a typo for $\beta+t\alpha$; I do admit though, I have always mixed up what is $\alpha$ and $\beta$ in the $\alpha$-string of $\beta$, and I'm sure you and I will see it mixed up at some point again.

    As for the one-dimensionality of $[\mathfrak g_\alpha, g_{-\alpha}]$, this would follow from the one-dimensionality of the $\mathfrak g_\alpha$; unfortunately, I see he only proves this afterwards ("Proposition C"), and it seems like he uses the result somewhere on the way in Prop. B. So: Good question.

     – Torsten Schoeneberg Nov 14 '22 at 21:50
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    One can show one-dimensionality of the root spaces $\mathfrak{g}_\alpha$ without a circular argument, see e.g. end of answer to https://math.stackexchange.com/q/3135978/96384. – Torsten Schoeneberg Nov 14 '22 at 21:56
  • @TorstenSchoeneberg So are you saying there that each string gives rise to a (simple?) representation of $\mathcal{sl}_2$, and 1-dimensionality follows from that? – wlad Nov 14 '22 at 22:24
  • @TorstenSchoeneberg By the way, thank you for the remarks so far. They're very helpful. – wlad Nov 14 '22 at 22:35
  • It is clear that each string gives rise to an $sl_2$-representation. But I think the argument in the link goes a bit differently: First, one shows each $L_\alpha$ is one-dimensional by considering the representation $\sum_{j\in \mathbb Z} L_{j \alpha}$ and using that $span(X)= [L_0, X] = L_\alpha$ via $sl_2$-representation theory and definition of $X$. Then, using that (the OP in the question claimed they already knew this one-dimensionality), one can show that each other string (i.e. those that do not go through $0$) is indeed a simple $sl_2$-rep. – Torsten Schoeneberg Nov 15 '22 at 00:31
  • Note that $\sum_{j \in \mathbb Z} L_{j \alpha}$ turns out to be $L_{-\alpha} \oplus L_0 \oplus L_\alpha$ and hence is (unless $L = sl_2$) actually not simple, as $L_0$ has in general higher dimension. That's why in that case, Samelson defines the root string differently at the end, replacing $L_0$ with just the one-dimensional space spanned by $h_\alpha$. Then it is irreducible (actually, literally isomorphic to $sl_2$ itself). But to get there, one might have to first consider the string with the "thickened" $L_0$ in the middle, and use the "Corollary 2" of the linked answer on that. – Torsten Schoeneberg Nov 15 '22 at 00:35

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