What is the Lie group smooth structure of quaternions?

It's obvious that Quaternions, (denote by $H$, without $0$) form a non-commutative group under multiplication ( it's even non commutative division algebra ). It seems that it's also obvious that Quaternions is a Lie Group, but somehow I don't understand what is the smooth map for $\mu: H \times H \rightarrow H$. Is the smooth map just a product of two quaternions, like $\mu(g,q)=gq$? Can someone help me on this?

Solution 1:

Note that, for any Lie group, the product map $G \times G \to G$ is required to be smooth (it's not enough for some random map $G \times G\to G$ to be smooth; rather, the product map that you wrote down must be smooth).

One way to see that this holds for the non-zero quaternions is to embed the quaterinons in $GL_2(\mathbb{C})$. The multiplication map on $GL_2(\mathbb{C})$ is obviously smooth (it's a polynomial in the entries) and so its restriction to $H^\times$ is also smooth.

You could also check it directly. The non-zero quaternions look like $\mathbb{R}^4$ with an origin removed, so inherit a smooth structure from that. Again, the multiplication law is just a polynomial in the coefficients of $1, i, j, k$ with respect to this identification, and so is smooth.

Actually the quaternions are even a ring-object in the category of differential manifolds -- both the addition and multiplication structures are smooth.