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As far as I am aware, for a group to be the direct product of groups it is sufficient that $G\cong H\times K$ as a set and $H$ and $K$ are normal in $G$. For Lie groups, normal groups corresspond to ideals of the Lie algebra. For $G$ compact, given an ideal $\mathfrak{h} \subset \mathfrak{g}$, we can always find a complementary ideal $\mathfrak{k}$ such that $\mathfrak{g}\cong\mathfrak{k}\oplus\mathfrak{h}$ by using an appropriate inner product, however, in order for the the group to be a direct product proofs require that $G$ be simply connected. As it seems we can get the product structure on $G$ from using the fact that $exp$ is surjective, I am not sure where the requirement for $G$ to be simply connected comes in... Additionally, if $G$ is not simply connected, is there a condition one can add to ensure the group is still a direct product?

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First, your original supposition is wrong: If we let $G = \mathbb{Z}_4$ and $H = K = \mathbb{Z}_2$, then $G\cong H\times K$ as sets (since both are $4$ element sets), but $G$ is not a non-trivial direct product.

In the Lie group setting, it makes more sense to ask that $G\cong H\times K$ as smooth manifolds. There are still counterexamples:

Let $G' = S^1\times SU(2n)$ and $G = G'/\pm(1,I)$. We let $H$ be the image of the $S^1$ factor in $G'$ in $G$, and $K$ be the image of the $SU(2n)$ factor.

Because $H$ and $K$ are images of normal subgroups under a surjective map, they are normal subgroups of $G$. Further, we claim that $G$ is diffeomorphic to $H\times K$.

To make this diffeomorphism, we first let $\phi:S^1\rightarrow SU(2n)$ be given by $\phi(z) = \operatorname{diag}(z,z,z...,z, \overline{z}, \overline{z},...,\overline{z})$, where there are $n$ copies of $z$ and $\overline{z}$. Then $\phi$ is a homomorphism, embedding $S^1$ into $SU(2n)$.

Now, let $f:G'\rightarrow H\times K$ be given by $f([z,A]) = (z^2, \phi(z)A)$.

This is well defined: $f([-z,-A]) = (z^2, \phi(z)A) = f([z,A])$.

This is injective: if $(z^2, zA) = (w^2, wB)$, then $z^2 = w^2$ which forces $z = \pm w$. If $z = w$, clearly $A=B$. If $z = -w$, clearly $A = -B$, so $(w,B) = -(z,A)$, so $[w,B] = [z,A]$.

This is surjective: Given $(w,B)\in H\times K$, let $z$ be any square root of $w$. Then $f([z, \phi(z^{-1})B]) = (w,B)$.

This is smooth: If one precomposes with the map $G'\rightarrow G\rightarrow H\times K$, one gets the map $(z,A)\mapsto (z^2, zA)$, which is obviously smooth. Since $G'\rightarrow G$ is a submersion, $f$ must be smooth as well.

The inverse is smooth: At a point $(w,B)\in H\times K$, we pick a square root of $w$ and call it $z$. Then, locally, the inverse of $f$ at $(w,B)$ looks like $([\sqrt{w}, \sqrt{w}^{-1} B])$. Since the square root function, viewed as a map $\mathbb{C}\setminus\{0\}\rightarrow \mathbb{C}$ is locally smooth (if we keep picking the same branch), this is smooth.

As far as your intuition regarding $\exp$, the issue is that we could have $\exp(h) = \exp(k)\neq e$ for some $h\in \mathfrak{h}\subseteq \mathfrak{g}$ and $k\in\mathfrak{k}\subseteq \mathfrak{g}$. For the $G$ example above, the vector $h = \pi \in \mathfrak{h}\oplus\mathfrak{k}$ and $k = \pi I$ exponentiate to $(-1, I)$ and $(1,-I)$ in $H\times K$. But in $G$, these get identified to the same (non-identity) point.

Finally, to answer your last question:

Suppose $G$ is any compact Lie group for which $Z(G) = \{e\}$, i.e., $G$ is centerless. If $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{k}$ with both $\mathfrak{h}$ and $\mathfrak{k}$ ideals, then $G$ is a direct product.

(So, for example, $SO(4)/\pm I$ must be a direct product, and indeed, it is $SO(3)\times SO(3)$.)

Proof: Let $\pi: G'\rightarrow G $ denote the universal cover of $G$. Using the group theoretic fact that $\pi(Z(G'))\subseteq Z(G) = \{e\}$, it follows that $Z(G')\subseteq \ker \pi$. The reverse inclusion $\ker \pi \subseteq Z(G')$ is always true for the universal cover (see this answer by Jack Lee and the references he gives). Thus, $Z(G') = \ker \pi$. Of course, writing $G' = H\times K$ (where $H$ and $K$ are the unique simply connected groups with Lie algebras $\mathfrak{h}$ and $\mathfrak{k}$), it follows that $Z(G') = Z(H)\times Z(K)$.

Then $G = G'/\ker \pi = (H\times K)/(Z(H)\times Z(K)) = (H/Z(H)) \times (K/Z(K))$ is a direct product, as claimed. $\square$

Edit Your claim in the comment is correct:

Theorem Assume $G$ is a connected compact Lie group with Lie algebra $\mathfrak{g}$, which decomposes as a sum of two ideals $\mathfrak{h}\oplus \mathfrak{k}$. Assume further that $H:=\exp(\mathfrak{h})$ and $K:=\exp(\mathfrak{k})$ intersect only at $e\in G$. Then $G$ is Lie isomorphic to $H\times K$.

Proof: Consider $f:H\times K\rightarrow G$ given by $f(h,k) = hk$. This is clearly smooth because the multiplication in $G$ is smooth. This map induces an isomorphism on the Lie algebra level, so it follows from general Lie theory that $f$ is a covering.

However, $f$ is injective: if $f(h,k) = e$, then $hk = e$, so $h = k^{-1} \in H\cap K =\{e\}$. An injective covering map is a diffeomorphism, so we are done.

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  • $\begingroup$ Thank you for such a thorough and informative answer! $\endgroup$ – R Mary Jul 24 '18 at 15:31
  • $\begingroup$ Dear Jason, sorry to bother you again after you put such trouble into your answer, but just so I know I have fully understood: the problem here is that even though the lie algebra split $\mathfrak{g}\cong\mathfrak{k}\oplus\mathfrak{h}$, we don't necessarily have $exp(\mathfrak{k})\cap exp(\mathfrak{h})=\{e\}$. So if $G$ is not simply connected, checking this would be enough i.e. the statement "if the lie algebra of $G$ is the sum ideals $\mathfrak{h}, \mathfrak{k}$ and $exp(\mathfrak{k})\cap exp(\mathfrak{h})={e}$, then $G$ is the direct product of Lie groups" is true? $\endgroup$ – R Mary Jul 24 '18 at 17:47
  • $\begingroup$ I am not certain, but I think that's right. That's certainly the condition you need on the group level to guarantee that that $G$ is isomorphic to $H\times K$. (That is, for an abstract group $G$, if you have normal subgroups $H,K\subseteq G$ with $H\cap K = \{e\}$, then $G\cong H\times K$). I am not 100% confident about the topological aspects, but I would bet everything works. I'll think about it for a bit and then edit the answer if I can get all the details straight. $\endgroup$ – Jason DeVito Jul 24 '18 at 17:53

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