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I wondered whether there is an analogue of the Krull-Schmidt theorem for real Lie-groups. More precisely, what conditions do you have to impose on a connected finite dimensional Lie group $G$ so that for each direct product decomposition $G=G_1 \times ... \times G_s$ the isomorphism classes of $G_1,...,G_s$ are uniquely determined up to reordering?

I'm mainly interested in compact Lie groups, maybe this restriction simplifies the matter.

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    $\begingroup$ I guess you mean connected? Because otherwise, all discrete groups are real Lie groups... $\endgroup$ – YCor Jan 27 '15 at 22:04
  • $\begingroup$ @YCor: Of course. $\endgroup$ – Dominik Jan 28 '15 at 15:06
  • $\begingroup$ Then each direct summand is a connected Lie subgroup, and the chains of connected closed subgroups have bounded length (bounded by the dimension). Is this enough to make the proof of Krull-Schmidt work? $\endgroup$ – YCor Jan 28 '15 at 15:47
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I see that you are restricting to (connected) compact Lie groups. This means that $G$ is a product of a semi-simple group and a central torus. The semi-simple piece decomposes uniquely (up to isomorphism) into a product of normal subgroups and the torus decomposes $T^n=S^1 \times \ldots \times S^1$. Thus

\begin{equation} G=G_{ss_1}\times\ldots \times G_{ss_k}\times S^1 \times \ldots S^1 \end{equation}

Where the first $k$ terms are semi-simple. If you demand that the factors are indecomposable and that they are Lie subgroups, ie no irrational winding on the torus, then this decomposition is unique. I hope this is what you wanted.

I have also been thinking about the non-compact case. I'm going to sketch my thoughts so we can get some discussion going, but there will be mistakes and it is unfinished. If G is semi-simple or abelian we get the same decomposition as before. Suppose $G$ is not semi-simple or solvable. Levi decomposition does give us something, but the Levi factor $G_{ss}$ is not normal in general, and it might even be non-closed which would probably stop us right here. On the Lie algebra level, we can split off the kernel $\mathfrak{g}_{ss_0}=\text{Ker}(\varphi)$ of the $\mathfrak{g}_{ss}$-representation $\varphi:\mathfrak{g}_{ss}\rightarrow \text{End}(\mathfrak{r})$, where $\mathfrak{r}$ is the solvable radical of $\mathfrak{g}$. $\mathfrak{g}_{ss_0}$ decomposes uniquely as it is semi-simple.

I don't know what to do with the complement to $\mathfrak{g}_{ss_0}$, and describing the unique decomposition directly seems hard. The approach I was considering was to use that $\mathfrak{g}_{ss}$-representations decompose into irreducibles, that an ideal of the radical is necessarily a submodule, and possibly split up $\mathfrak{g}_{ss}$ further by taking the kernels of sub-representations. In the end this didn't seem to help much.

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  • $\begingroup$ It is true that any connected compact Lie group decomposes uniquely into a product of a semisimple group and a central torus. However, this product does not have to be direct. For example, $SU(n)$ is not a direct product of a semisimple group and a torus. $\endgroup$ – Mikhail Borovoi Feb 1 '15 at 21:15
  • $\begingroup$ @MikhailBorovoi I would call $SU(n)$ semi-simple. It does not need to decompose further. There is no torus in this case. $\endgroup$ – Henrik Winther Feb 2 '15 at 8:38
  • $\begingroup$ Excuse me, I meant that $U(n)$ is not a direct product of a semisimple group and a torus. $\endgroup$ – Mikhail Borovoi Feb 3 '15 at 9:54
  • $\begingroup$ @MikhailBorovoi I see the problem now. $U(n)$ is only covered by $SU(n)\times S^1$, not equal to. If I find the time I will edit the answer to account for this situation. Thank you. $\endgroup$ – Henrik Winther Feb 5 '15 at 13:53

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