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?


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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.