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The adjoint representation of $\mathfrak{sl}_2(\mathbb{C})$ under a natural basis, it is given by $$\text{ad}: \mathfrak{sl}_2(\mathbb{C})\to\mathfrak{gl}_3(\mathbb{C})$$ $$\left(\begin{matrix}a&b\\c&-a\end{matrix}\right)\mapsto \left(\begin{matrix}0&-c&b\\-2b&2a&0\\2c&0&-2a\end{matrix}\right).$$ We see that this homomorphism of Lie algebras is injective and its image really looks like $\mathfrak{o}_3(\mathbb{C})$, except the $2$'s.

So I wonder if with a good choice of basis I can get an isomorphism between the Lie algebras $\mathfrak{o}_3(\mathbb{C})$ and $\mathfrak{sl}_2(\mathbb{C})$ (if they are isomorphic)... I've been trying some, but I did not succeed. Is there an intelligent way to see if such a good basis exists or not ?

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How did you obtain this? There should be zeroes on the diagonal (in the standard basis) since we have $ad_x(x) = [x,x] = 0$. Can you elaborate which basis in $\mathfrak sl_2(\mathbb C)$ you use? –  Marek Jan 12 '13 at 16:31
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@Marek No, the calculation is correct. Even if $[x,x]=0$, there is no reason to have zeroes on the diagonal. –  Klaus Jan 12 '13 at 16:37
    
My bad, it's really correct. –  Marek Jan 12 '13 at 16:48
    
The terminology in this question is confusing. When you say $ad : sl_2 \to gl_3$ what you mean to say is << Given $X = \left(\begin{array}{cc} a & b \\ c & -a \end{array}\right)$ we identify the endomorphism $ad_X$ of $\mathfrak{sl}_2$ with a matrix in $\mathfrak{gl}_3$ upon identifying the basis vectors of $\mathfrak{sl}_2$ with the standard basis of $\Bbb{C}^3$.>> –  user38268 Jan 12 '13 at 17:00
    
I didn't use the same terminology, but I ended up with the exact same computations... am I lucky :P good question –  Patrick Da Silva Jun 14 '13 at 8:35

2 Answers 2

up vote 3 down vote accepted

$\newcommand\ad{\operatorname{ad}}$Let $\beta$ be the Killing form on $\mathfrak{sl}_2$, so that for $X$, $Y\in \mathfrak{sl}_2$ we have $$\beta(X,Y)=\operatorname{tr} \ad(X)\circ \ad(Y).$$ You can easily check that this is a non-degenerate symmetric bilinear form on $\mathfrak{sl}_2$. Moreover, the adjoint action of $\mathfrak{sl}_2$ on itself respects this, in the sense that $$\beta(\ad(X)(Y),Z)+\beta(Y,\ad(X)(Z))=0$$ for all $X$, $Y$, $Z\in sl_2$. If we let $\mathfrak{o}(\beta)\subseteq \mathfrak{gl}(sl_2)$ be the Lie algebra of endomorphisms of the vector space $\mathfrak{sl}_2$ which respect the bilinear form $\beta$, then this tells us that the image of $\ad:\mathfrak{sl}_2\to \mathfrak{gl}(\mathfrak{sl} 2)$ is contained in $\mathfrak{o}(\beta)$.

Now check that $\mathfrak{o}(\beta)\cong \mathfrak{o}_3$. Basically, this is because if $b$ and $b'$ are two bilinear forms on a vector space $V$, then $\mathfrak{o}(b)\cong\mathfrak{o}(b')$ whenever there is an isomorphism of quadratic spaces $(V,b)\cong(V,b')$. And all non-degenerate symmetric bilinear forms on a complex vector space are isomorphic.

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Thank you very much, very interesting! –  Klaus Jan 12 '13 at 16:43
    
Don't accept this answer until you have checked all I wrote! :-) –  Mariano Suárez-Alvarez Jan 12 '13 at 16:50

Calculate the dimensions of both Lie algebras. Once you've done that, see if you can find the image of the map by finding an extra relation satisfied by the matrix on the right not satisfied by an arbitrary element in $\mathfrak{o}_3$.

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