# What is the manifold structure of U(n)?

A Lie group is simultaneously a differentiable manifold. As I understand it, the Lie group is generated via exponentiation of the generators of the Lie algebra.

It is intuitively clear to me that the Lie group U(1) is isomorphic to the manifold of the circle. Are SU(2) and SO(3) isomorphic to the two-sphere?

Furthermore, what is the manifold structure of U(n) in general? Furthermore, can anything be said about the manifold structure of the infinite-dimensional Unitary group? (This group occurs all the time in applications in quantum physics when the Hilbert space is infinite-dimensional).

• – Joe Johnson 126 Jun 22 '14 at 19:11
• "The Lie group is generated via exponentiation of the generators of the Lie algebra" is not precisely true. In general, the connected component of the identity element is generated by exponentials. If the Lie group is also compact, the exponential map is surjective. (All of the examples you mention are connected and compact, but e.g. $O(3)$ is not connected.) – user111187 Jun 22 '14 at 19:18

$\text{SU}(2)$ is in fact diffeomorphic to the $3$-sphere $S^3$, rather than the $2$-sphere (which has the wrong dimension; recall that $\mathfrak{su}(2)$ is $3$-dimensional); this comes from its identification with the unit quaternions. $\text{SO}(3)$ is diffeomorphic to the real projective space $\mathbb{RP}^3$, which is the quotient of $S^3$ obtained by identifying antipodes.

In general, $\text{U}(n)$ can't be identified as a more familiar-looking manifold. It is an "iterated extension" of the odd-dimensional spheres $S^1, S^3, ..., S^{2n-1}$, and in fact is rationally homotopy equivalent to the product $S^1 \times S^3 \times ... \times S^{2n-1}$. This means in particular that it has the same rational cohomology and rational homotopy groups as this product; however, it is in general not not homeomorphic or diffeomorphic to this product. "Iterated extension" means that the unitary groups fit into fiber sequences which are ultimately built from odd spheres, starting with

$$\text{SU}(n) \to \text{U}(n) \xrightarrow{\det} S^1$$

and continuing with

$$\text{SU}(n-1) \to \text{SU}(n) \to S^{2n-1}.$$

The first sequence is even a short exact sequence of Lie groups, and it splits smoothly, so $\text{U}(n)$ is diffeomorphic to $\text{SU}(n) \times S^1$; in particular $\text{U}(2)$ is diffeomorphic to $S^3 \times S^1$. But this is not an isomorphism of groups ($\text{U}(n)$ is a semidirect product rather than a direct product), and the corresponding statement for the other fiber sequences should be false, although I haven't verified this. I do know that $\text{SU}(3)$ is not diffeomorphic, and in fact is not even homotopy equivalent, to $S^3 \times S^5$; see this MO question.

The infinite-dimensional unitary group is not a manifold in the usual sense, although it is a Hilbert manifold. Kuiper's theorem implies that it is weakly contractible, so from the perspective of homotopy theory it looks like a point. But of course what actually appears in physics is not the unitary group but the projective unitary group; the infinite-dimensional projective unitary group is an Eilenberg-MacLane space $K(\mathbb{Z}, 2)$. This comes from its identification as a quotient of a weakly contractible space, namely the infinite-dimensional unitary group, by a free action of $\text{U}(1)$; hence it is a model for the classifying space $B \text{U}(1)$, and $\text{U}(1) \cong S^1$ itself is a $K(\mathbb{Z}, 1)$.

A closely related space called the stable unitary group is also very interesting from the perspective of homotopy theory; its homotopy groups are $2$-periodic, which is one way of stating complex Bott periodicity.