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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?

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

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    $\begingroup$ I think you've already basically hit on this, but the unit quaternions are diffeomorphic to $SU_2$ and they are also isomorphic as Lie groups. Since multiplication on $SU_2$ is smooth so too must multiplication on the unit quaternions be. Furthermore the quaternions admit the polar decomposition $q = re^{\Theta\mu}$, where $\mu$ is pure and $r$ is real, thus we construct $H$ as a union of nested hyperspheres to turn the unit quaternion / $SU_2$ Lie group isomorphism into an injective ring homomorphism from $H\rightarrow GL_2(\mathbb{C})$. $\endgroup$ – Mortified Through Math Jun 1 '16 at 1:38

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