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I've started Lie algebras course this semester, so I already have some questions. Well, I need to find a $Lie(SL_n(\mathbb{K}))$ (Lie algebra). Field $\mathbb{K}$ is algebraically closed and $char( \mathbb{K})=0$. Teacher gave me next piece of advice. This space is given by the equation $detg=1,\;g\in GL_n$. We have a formula for determinant of matrix $g$.

$$detg=\sum_{\sigma \in S_n}sgn(\sigma)a_{1\sigma (1)}...a_{n\sigma (n)}$$

So I need to differentiate this identity at E. But I thought that $a_{ij}$ are elements of out field, so $d(a_{ij})=0$. Anyway I know, that I need to get something like this:

$$d(detg)=\sum_{i=1}^nda_{ii}=0$$

So the Lie algebra of $SL_n$ is subspace of matrices with zero trace. Please, help me understand the differentiating part.

Thanks.

Kirill Losev
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  • As I understand your confusion - isn't it similar to saying that if $f(x)= x$, then $f'(1) = 0$, because $1'=0$? – peter a g Sep 17 '18 at 17:13
  • @peterag If i understand you right then yes. I understand why there is a zero on the right side. But why do we have trace on the left side? – Kirill Losev Sep 17 '18 at 17:17
  • Oh, I get your idea. So $a_{ij}$ not just numbers, but some kind of variables. Like coordinates of $n^2$ dimensional space. So when I differentiate it I get the value of $a_{ij}$ variable. – Kirill Losev Sep 17 '18 at 17:27
  • Yes, you are computing the degree-1 part of $\det(I+A)=1$ as $A\in\mathfrak{gl}_n$, when you regard $\det$ as a polynomial function in $n^2$ variables. – user10354138 Sep 17 '18 at 17:31
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    Right. The Lie group version would be to consider a path $t\to \gamma(t)$ in $SL_n({\mathbb R})$ with $\gamma (0) =I$. Then $\det \gamma (t) = 1$, and one would differentiate on both sides. To make this practical, write $\gamma (t ) = 1 + t X + t^2\cdot\text{something}$, so $\det (1 + t X + t^2\cdot\text{something}) =1 $. Use the formula for $\det$ to see that the LHS $1 + t \text{Tr} X + t^2\cdot\text{something}$, and differentiate. (cont) – peter a g Sep 17 '18 at 17:33
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    (cont) In the language of algebraic geometry, 'I' would look at matrices M with ceoff's in the 'dual numbers' $K[\epsilon]$, where $\epsilon^2 = 0$, such that $M = I + e X$. – peter a g Sep 17 '18 at 17:33
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    And, of course, in the Lie group version, we are interested in $\gamma' (0) = X$; in the alg. geo version $X$ has coefficients in $K$. – peter a g Sep 17 '18 at 17:40
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    See also this question. – Dietrich Burde Sep 17 '18 at 21:26

1 Answers1

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Well I'll try to answer my question on my own. Let's differentiate first sum. $$ d(detg)=\sum_{\sigma \in S_n} (-1)^\sigma\sum_{i=1}^nd(a_{i\sigma(i)})a_{1\sigma(1)}...\hat{a_{i\sigma(i)}}...a_{n\sigma(n)}=0$$ Now lets restrict it to the E. Members of this sum are not zero iff $\sigma=id$.

$$d(detg)\vert_E=\sum_{i=1}^nd(a_{ii})=0$$

Kirill Losev
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