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I have an exercise which asks to prove that the Lie algebra $\mathfrak{sl}_2(\mathbb{C})$ contains the real, non-isomorphic subalgebras $\mathfrak{sl}_2(\mathbb{R})$ and $\mathbb{R}^3$ and to show further that - as vector spaces - each of these two subalgebras spans $\mathfrak{sl}_2(\mathbb{C})$ over $\mathbb{C}$.

I'm a bit confused by this statement. Isn't a subalgebra of a Lie algebra simply a linear subspace $\mathfrak{h} \subseteq \mathfrak{g}$ such that $[\mathfrak{h},\mathfrak{h}] \subseteq \mathfrak{h}$? If yes, clearly multiplying an element of $\mathfrak{sl}_2(\mathbb{R})$ by a real number yields again an element of $\mathfrak{sl}_2(\mathbb{R})$, and the subspace is closed under the commutator too. On the other hand, I don't understand how to view $\mathbb{R}^3$ as a subspace of $\mathfrak{sl}_2(\mathbb{C})$.

I'm guessing I'm misunderstanding a crucial part of this exercise. Could anyone explain what is actually asked or how to view $\mathbb{R}^3$ as a subspace of $\mathfrak{sl}_2(\mathbb{C})$?

user302234
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2 Answers2

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Identify $\mathbb{C}$ with $\mathbb{R}^2$ and extend the monomorphism $x\mapsto (x,0)$ from $\mathbb{R}\hookrightarrow \mathbb{R}^2=\mathbb{C}$ to a monomorphism $\mathfrak{sl}_2(\mathbb{R})\hookrightarrow \mathfrak{sl}_2(\mathbb{C})$. This is an injective Lie algebra homomorphism. The Lie algebra $\mathfrak{sl}_2(\mathbb{R})$ has an abelian subalgebra of dimension $1$, but not of dimension $2$. Hence the maximal dimension of an abelian (real) subalgebra of $\mathfrak{sl}_2(\mathbb{C})$ is equal to $2$. Hence there is no faithful morphism from $\mathbb{R}^3\hookrightarrow \mathfrak{sl}_2(\mathbb{C})$. As real vector spaces, $\dim \mathfrak{sl}_2(\mathbb{R})=\dim \mathbb{R}^3=3$ and $\dim \mathfrak{sl}_2(\mathbb{C})=6$. As vector space, consider $\mathfrak{sl}_2(\mathbb{R})$ as $3$-dimensional subspace of $\mathbb{R}^4=M_2(\mathbb{R})$, which is $\mathfrak{gl}_2(\mathbb{R})$ under the Lie bracket $[A,B]:=AB-BA$.

Dietrich Burde
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  • Thank you for your answer. What I don't understand in both your and Julio Cesar's answer: Why are you arguing that there can not be an abelian real subalgebra? Isn't the usual Lie bracket on $\mathbb{R}^3$ the cross product - which is clearly not abelian? Maybe it is possible to find an injective Lie algebra homomorphism $(\mathbb{R}^3, \times) \to (\mathfrak{sl}_2(\mathbb{C}), [ \cdot, \cdot])$? – user302234 Jan 06 '16 at 14:35
  • Sure there can be a non-abelian Lie algebra, namely $\mathfrak{sl}_2(\mathbb{R})$ itself. But not $\mathbb{R}^3$ with bracket $[A,B]=0$ inside the matrix Lie algebra $\mathfrak{sl}_2(\mathbb{C})$. Or can you give me three linear independent matrices with trace zero which have pairwise zero commutator ? By the way, $(\mathbb{R}^3,\times)$ is the Lie algebra $\mathfrak{so}_3(\mathbb{R})$. – Dietrich Burde Jan 06 '16 at 15:57
  • I've thought about this for a while and carefully checked Julio Cesar's answer, which clearly gives an isomorphism $(\mathbb{R}^3, \times) \cong \mathfrak{su}_2(\mathbb{C})$ which also is a Lie algebra homomorphism. Also, $\mathfrak{su}_2(\mathbb{C})$ is clearly a Lie subalgebra of $\mathfrak{sl}_2(\mathbb{C})$. Is there something wrong with his argument that I'm missing? I did the calculations and can't find out what's wrong, but it clearly contradicts your statement. – user302234 Jan 09 '16 at 12:48
  • It is also immediately clear that - as a vector space - $\mathfrak{su}_2(\mathbb{C})$ spans $\mathfrak{sl}_2(\mathbb{C})$ over $\mathbb{C}$... – user302234 Jan 09 '16 at 12:59
  • Why do you see a contradiction ? The Lie algebra $\mathbb{R}^3$ is abelian, whereas $(\mathbb{R}^3,\times)$ is a totally different (non-abelian) Lie algebra, which is not asked for in your exercise. – Dietrich Burde Jan 09 '16 at 13:18
  • What I meant by my very first comment was: "Maybe by the Lie algebra $\mathbb{R}^3$ the author of the exercise actually means the Lie algebra $(\mathbb{R}^3, \times)$, which then would indeed be realizable as a subalgebra of $\mathfrak{sl}_2(\mathbb{C})$, right?". If you agree, then I am happy, because it seemed to me like you had disagreed and that's why I saw a contradiction. – user302234 Jan 09 '16 at 13:21
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    Maybe. But the people seeing your title question will certainly think that $\mathbb{R}^3$ is the abelian subalgebra, whereas $(\mathbb{R}^3,\times)$ is the simple Lie algebra $\mathfrak{so}_3(\mathbb{\mathbb{R}})$. – Dietrich Burde Jan 09 '16 at 13:25
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    So the title should read, in that case, "$\mathfrak{sl}_2(\mathbb{R})$ and $\mathfrak{so}_3(\mathbb{R})$ as subalgebras of $\mathfrak{sl}_2(\mathbb{C})$". This would really make sense. It would be a different question, of course. – Dietrich Burde Jan 09 '16 at 14:00
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You won't find any $\mathbb R$-subspace $\mathfrak h$ of $\mathfrak{sl}_2(\mathbb C)$ such that $[\mathfrak h,\mathfrak h]=0$ whose real dimension is $3$. In fact, you would have that $\mathfrak h_\mathbb C=\mathfrak h\oplus i\mathfrak h$ is a complex abelian sub-algebra of $\mathfrak{sl}_2(\mathbb C)$ whose complex dimension is $2$ or $3$. That means $$ \mathfrak{sl}_2(\mathbb C)=\mathfrak h_\mathbb C\oplus V, $$ where $V$ is a complex $1$-dimensional subspace, or $$ \mathfrak{sl}_2(\mathbb C)=\mathfrak h_\mathbb C. $$ So, in the firs case, $$ \mathfrak{sl}_2(\mathbb C) = [ \mathfrak{sl}_2(\mathbb C), \mathfrak{sl}_2(\mathbb C) ] = [\mathfrak h_\mathbb C,\mathfrak h_\mathbb C]\oplus[\mathfrak h_\mathbb C,V]\oplus[V,V]=[\mathfrak h_\mathbb C,V]. $$ But $\mathfrak{sl}_2(\mathbb C)$ has complex dimension $3$ and $[\mathfrak h_\mathbb C,V]$ has complex dimension at most $2$. A contradiction. In the second case, $$ \mathfrak{sl}_2(\mathbb C) = [ \mathfrak{sl}_2(\mathbb C), \mathfrak{sl}_2(\mathbb C) ] = [\mathfrak h_\mathbb C,\mathfrak h_\mathbb C]=0. $$ Again, a contradiction.

So, you can view the $3$-dimensional abelian Lie algebra $\mathbb R^3$ as a $\mathbb R$-subspace of $\mathfrak{sl}_2(\mathbb C)$ but not as a sub-algebra of it.

From what you wrote, my educated guess about it would be that the exercise is really about you noting that:

  • $\mathfrak{sl}_2(\mathbb R)$ can be canonically included in $\mathfrak{sl}_2(\mathbb C)$;

  • $\mathbb R^3$ can NOT be included (or viewed as OP said) in $\mathfrak{sl}_2(\mathbb C)$;

  • $\mathfrak{sl}_2(\mathbb R)$ and $\mathbb R^3$ span different complex Lie algebras.

Edited after OP's comment:

Consider $\mathfrak{su}(2)$ the real Lie algebra of the $2\times 2$ traceless antihermitian matrices: $$ \left\{ \left( \begin{array}{cc} i\alpha & \beta + i\gamma \\ - \beta + i\gamma & -i\alpha \end{array} \right) : \alpha,\beta,\gamma\in\mathbb R \right\}. $$ You can check directly (if you haven't yet) that $\mathfrak{su}(2)$ is a real Lie algebra (with the matrix commutator as the Lie bracket). Also, you can realize it as a real subalgebra of $\mathfrak{sl}_2(\mathbb C)$.

The Lie algebra $(\mathbb R^3,\times)$ (where $\times$ denotes the vector product in $\mathbb R^3$) is isomorphic to $\mathfrak{su}(2)$ by the map: $$ (x,y,z)\to\frac{1}{2}\left( \begin{array}{cc} ix & y + iz \\ - y + iz & -ix \end{array} \right). $$

So, this is one of the ways you can realize $(\mathbb R^3,\times)$ as a real sub-algebra of $\mathfrak{sl}_2(\mathbb C)$.

Júlio César
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  • You have implicitly assumed that $\mathfrak{h} \cap \mathfrak{ih} = { 0 }$ but there is no reason this has to hold; see the other answer. – hunter Jan 05 '16 at 22:34
  • @hunter You're right! I edited my answer considering the case in which $\mathfrak h\cap i\mathfrak h\neq 0$. – Júlio César Jan 05 '16 at 23:01
  • Thank you for your answer. What I don't understand in both your and Dietrich Burde's answer: Why are you arguing that there can not be an abelian real subalgebra? Isn't the usual Lie bracket on $\mathbb{R}^3$ the cross product - which is clearly not abelian? Maybe it is possible to find an injective Lie algebra homomorphism $(\mathbb{R}^3, \times) \to (\mathfrak{sl}_2(\mathbb{C}), [ \cdot, \cdot])$? – user302234 Jan 06 '16 at 14:37
  • @user302234 Aaaaw right! That might be it! The Lie algebra $(\mathbb R^3,\times)$ is not abelian. I will edit my answer. – Júlio César Jan 06 '16 at 18:27