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Show that $\mathfrak{su}(n)$ and $\mathfrak{sl}(n,\Bbb R)$ are not isomorphic as real Lie algebras.

I calculated that both of them have the Killing form $$K(x,y)=2n \,\textrm{tr}(xy) .$$ I got the hint that I should consider the signature of both Killing forms. Can you tell me how I can use this hint?

Travis Willse
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user95
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    What's the question exactly? Have you computed the signatures of the Killing forms? – Travis Willse May 30 '17 at 15:44
  • I want to show that su(n) and sl(n,R) are not isomorphic. But I don't know how I can calculate the signature. Second I dont know why different signatures implie that these Lie algebras are not isomorphic. – user95 May 30 '17 at 15:52
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    The Killing form is natural, so if you have a Lie algebra isomorphism $\Phi: \frak g \to \frak h$, then the Killing forms are related by $K_{\frak g} = \Phi^* K_{\frak_h}$. In particular, the signatures of $K_{\frak g}$ and $K_{\frak h}$ must coincide. – Travis Willse May 30 '17 at 16:07
  • If you want to show that these Lie algebras are not isomorphic, you can just look for dimensions of subalgebras, too. – Dietrich Burde Feb 11 '21 at 14:56

2 Answers2

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One can also show that $\mathfrak{su}(n)$ and $\mathfrak{sl}(n,\Bbb R)$ are not isomorphic without using the Killing form. One algebra has subalgebras of a certain dimension, the other one does not - see the following question:

Showing the Lie Algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb{R})$ are not isomorphic.

Furthermore, one algebra is compact, the other one not (here one could use the Killing form, which is negative definite for one algebra, but not for the other).

Dietrich Burde
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Hint For $X \in \mathfrak{su}(n)$ we have $X = -{\bar X}^{\top}$ and so $$K(X, X) = 2n\, \textrm{tr}(X^2) = -2n\, \textrm{tr}(\bar X^{\top} X) .$$ Can you show that this quantity is nonpositive?

Travis Willse
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  • We have $tr(\bar{X}^TX)=\sum_{j=1}^n\sum_{i=1}^n \bar{x_{ij}}x_{ij}=\sum_{j=1}^n\sum_{i=1}^n |x_{ij}|^2\geq0$. Thus $K(X,X)\leq0$. But the Killingform of sl(2,R) is not negative definite and thus they are not isomorphic, right? – user95 May 30 '17 at 16:32
  • Looks good to me. Note that to show that the Killing form of $\frak{sl}(2, \Bbb R)$ is not negative definite, it is enough find some $X \in \frak{sl}(2, \Bbb R)$ such that $K_{\frak{sl}(2, \Bbb R)}(X, X) > 0$. – Travis Willse May 30 '17 at 17:51