The Lie algebras $su(2)$ and $sl(2,\mathbb C)$ have the same Dynkin Diagram (just a blob) and therefore also have the same structure constants and isomorphic Lie algebras. Additionally, they are both, as one can prove, simple and semisimple. But both Lie groups are not isomorphic (since the latter is non-compact) neither is one a covering group of the other. How is this possible?

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    $\begingroup$ First thing to note: $\mathfrak{su}(2)$ is a real Lie algebra, but is not a complex Lie algebra (in particular it is not a complex vector space!). On the other hand, $\mathfrak{sl}(2,\mathbb C)$ is naturally complex Lie algebra. The structure theorems, Dynkin diagrams etc. all assume that your field is algebraically closed. The point is that $\mathfrak{su}(2)$ and $\mathfrak {sl}(2, \mathbb R)$ are non-isomorphic Lie algebras, whose base changes to $\mathbb C$ are isomorphic. $\endgroup$ – Mathmo123 May 22 '18 at 18:16

They don't have the same Lie algebra : $SU_2$ has real dimension $3$ and $SL_2(\Bbb C)$ has complex dimension $3$. What is true is that $\Bbb C \otimes \mathfrak{su}_2 \cong \mathfrak{sl}_2$. To understand the connexion, keywords are "complexification of a Lie algebra" or "compact form of a complex Lie group".


Actually, also over the real numbers, the Lie algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}_2(\Bbb{R})$ are different; consider their the Killing forms:

The Killing forms of $\mathfrak{su}(n)$ and $\mathfrak{sl}(n,\Bbb R)$ are not isomorphic (as real Lie algebras)

In general, two (simple) Lie groups with the same Lie algebra can be different. For this topic, have a look at the posts at this site:

How are groups with the same Lie Algebra inequivalent?

Infinite number of Lie groups with the same lie algebra

Groups have the same Lie algebra


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