Semi-direct product and unitary group: need for subgroups to share only the identity.

Question

The Lie algebra $u(2)$, say, is the direct sum $u(1) \oplus su(2)$, where $u(1)$ has dimension one with generator in the fundamental which we may as well take to be $1\!\!1$. Yet the Lie group $U(2)$ is not isomorphic to the direct product of the Lie groups $U(1) \otimes SU(2)$ where the group elements in $U(1)$ are $e^{i t}1\!\!1$ for $t \in \mathbb{R}$.

Indeed, nor is this group a semi-direct product! The reason usually given is that although the elements of $U(1)$ commute with those of $SU(2)$ (or both are normal subgroups) both subgroups share more than the identity element. Indeed, they both share $-1\!\!1$. To see this, take $t = i\pi$ for $U(1)$ and $\exp(i \pi \hat{n}\cdot \sigma)$ where the Pauli matrices, $\sigma$, are (proportional to) the generators of $su(2)$.

My question is why this is a big deal - certainly in the construction of a direct product $G = G_{1} \otimes G_{2}$ one can decompose any element $(g_{1}, g_{2})$ into $(g_{1}, I_{2})(I_{1}, g_{2})$, a product of elements from the two subgroups. It is also clear that the subgroups $G_{1} = (g_{1}, I_{2})$ and $G_{2} = (I_{1}, g_{2})$ share only the identity, but I struggle to see why this convenience should necessarily be taken forward to the reverse case that one decomposes a group into a direct product.

In particular, for $U(2)$ the $U(1)$ and $SU(2)$ subgroups commute, any element can be decomposed into a product of elements in the two subgroups (even $(-1\!\!1, -1\!\!1) = (1\!\!1, -1\!\!1)(-1\!\!1, 1\!\!1)$) and the product of two elements $(g_{1}, g_{2})(g'_{1}, g'_{2}) = (g_{1}g'_{1}, g_{2}g'_{2})$ is the obvious product that direct product groups are endowed with. The only obstruction seems to be that they share two common elements and I would like to understand why this is a problem.

Thanks for the advice!

Answer 1

Let $U$ and $V$ be groups, and $Z(\_)$ denotes the center. The following is exercise $7.9$ in Isaac's algebra.

Let $M \subset Z(U)$ and $N \subset Z(V)$ and assume $M \cong N$. Then there is a group $G$ with normal subgroups $K$ and $L$ so that $K \cong U$ and $L \cong V$ and $KL = G$ and $[K,L] = 1$ (they commute) and $K \cap L = M$ (after identifying $K$ with $U$).

This is called the central product of $U$ and $V$ identifying $M$ and $N$.

Note that if you have two subgroup $G_1$ and $G_2$ of $G$, and you insist that they commute, then $G_1 \cap G_2$ is in the center of both of them.

See here for more information: https://en.wikipedia.org/wiki/Central_product

In particular, on that page an "internal central product" is described. It appears to be what you want.

(I don't know if this answers your question at all. But it would be an unreadable comment.)

Also, if you want to know why the "product" is so special, see this page: https://en.wikipedia.org/wiki/Product_(category_theory)

Answer 2

I think I've answered my own question - thank you to the community for offering useful suggestions.

If one wishes to try to decompose $G$ into two subgroups $G_{1}$ and $G_{2}$ and state that $G$ is isomorphic to the product group $G_{1} \otimes G_{2}$ with the usual product $(g_{1}, g_{2})(g'_{1}, g'_{2}) = (g_{1}g'_{1}, g_{2}g'_{2})$ then it must be the case that $G_{1} \cap G_{2} = 1\!\!1$ simply because every element of $G_{1} \otimes G_{2}$ can be written as $(g_{1}, g_{2}) = (g_{1}, 1\!\!1)(1\!\!1, g_{2})$.

Since $U(2)$ does not satisfy this requirement it cannot possibly be isomorphic to $U(1) \otimes SU(2)$ nor $U(1) \ltimes SU(2)$.