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This may be an unenlightening question, but I'm just not sure about the result and hoping someone can help me varify it.

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This question is related to these three questions.

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I want to construct the isomorphism relationship between the Lie Groups $SL(2,\mathbb{C})$ and $SU(2)$. I have the feeling that there should be some such isomorphism of groups.

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To begin, we know that as Lie Algebras

$$ \mathfrak{sl}(2,\mathbb{C}) \simeq \mathfrak{so}(1,3) $$

and

$$ \mathfrak{su}(2) \oplus \mathfrak{su}(2) \simeq \mathfrak{o}(4) $$

But we also know that

$$ \mathfrak{so}(n) \simeq \mathfrak{o}(n) $$

so I believe that this allows us to write

$$ \mathfrak{su}(2) \oplus \mathfrak{su}(2) \simeq \mathfrak{sl}(2,\mathbb{C}) $$

This makes sense anyway, since we know that the real algebra of the complexification of $\mathfrak{su}(2)$ is $\mathfrak{sl}(2,\mathbb{C})$, and in taking the real algebra of the complexified Lie algebra we get two commuting copies.

So, the part that I am not yet convinced about is how to get from this relationship between algebras to a relationship between groups.

I was told by someone in the department that

Theorem The Fundamental Theorem of lie Groups: Let $G_1$, $G_2$ be Lie groups. Then $G_1$ and $G_2$ have isomorphic Lie algebras if and only if they are locally isomorphic.

So this is a local statement only.

Moreover, he said that there is an extension of this theorem to a global statement which says that the Lie groups are globally isomorphic if they are simply connected.

Now, for our two groups, $SL(2,\mathbb{C})$ and $SU(2)$, we know that they are indeed simply connected. We could prove this, or instead, recall that they are the Universal Covering Groups of $SO(1,3)\uparrow$ and $SO(3)$ respectively, and so by the definition they must be simply connected.

This would solve our problem, and we could write down

$$ SU(2) \times SU(2) \simeq SL(2,\mathbb{C})$$

and be done.

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However I want to try to verify that statement, as opposed to taking it in blind faith (not that I have any reason to doubt it, but rather that I'd like to 'learn it' as opposed 'to be aware of it', if that makes sense).

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I tried looking it up, and the obvious source didn't have anything on a Fundamental Theorem of Lie Groups, only a short bit on The Third Theorem of Lie.

Some searching brought up these lecture notes (in .pdf format) from UCLA. It appears to be getting at what I want, but unfortunately is written in category theoretic language, which I know nothing about.

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Could anyone verify for me if this is correct, and perhaps point me to a book/ website/ lecture notes etc. where I could reference. (Our library is huge, so a book being online need not be a constraint).

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Comment to the question (v1): The two Lie groups $G_1=SL(2,\mathbb{C})$ and $G_2=SU(2)\times SU(2)$ are not isomorphic. For starters, $G_1$ is non-compact and $G_2$ is compact. –  Qmechanic May 21 at 14:25
    
Ah, ok, so then we certainly won't be able to find an isomorphism of groups between them! Is what I have written correct, if pointless, all the same? Namely, am I indeed correct in thinking that as Lie algebras $$ \mathfrak{su}(2) \oplus \mathfrak{su}(2) \simeq \mathfrak{sl}(2,\mathbb{C}) $$ ? Finally, does this mean that the best we can do is have a local isomorphism, as per that theorem of Lie? Thank you! –  Flint72 May 21 at 14:31
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The two real Lie algebras ${\rm Lie}(G_1)=sl(2,\mathbb{C})\cong so(1,3;\mathbb{R})$ and ${\rm Lie}(G_2)=su(2) \oplus su(2)\cong so(4;\mathbb{R})$ are not isomorphic. –  Qmechanic May 21 at 15:02
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Yes, the complexification is $sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$ for both Lie algebras. –  Qmechanic May 21 at 15:23
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SL(2, C) and SU(2) are not isomorphic, but there can be some homomorphism between the 2 –  Nikos M. May 22 at 23:28

1 Answer 1

You should be able to do it explicitly by mapping the Cartan subgroups of the SU's (diagonal matrices) to the Cartan subgroup of SL2 (also diagonal) and the root vectors to the root vectors. Do this concretely for the Lie algebras first. Then exponentiate to get it on the Lie Groups. Since the groups are simply connected, this will not lead to ambiguities.

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