# Question on universal property of Clifford algebra.

I have short question regarding the universal property of the Clifford algebra. Suppose $$(V,q)$$ is a quadratic $$\mathbb{F}$$-vector space and $$(\mathrm{Cl}(V,q),j)$$ is the Clifford algebra. Recall that $$\mathrm{Cl}(V,q)$$ is an associative and unital algebra and $$j:V\to \mathrm{Cl}(V,q)$$ is a linear map such that $$j(v)^{2}=-q(v)1_{\mathrm{Cl}(V,q)}$$. Then, the universal property is usually stated as follows:

For any other pair $$(A,j)$$ consiting of a unital associative algebra $$A$$ and a linear map $$k$$ such that $$k(v)^{2}=-q(v)1_{\mathrm{Cl}(V,q)}$$, there exists a unique algebra homomorphism $$\psi:\mathrm{Cl}(V,q)\to A$$ such that $$k\circ\psi=j$$.

Now to my question: Is the unique algebra homomorphism $$\psi$$ required to be unital? It seems a bit more natural to me, since the category we are working with is the one of unital associative algebras. However, in all the resources I saw (and I checked many of them), it was never written that the unique algebra homomorphism in the universly property is assumed to be unital. Is it implicitely understood? Is it actually redundant and follows from something else? Is it not required for some reason?

• Should be $j : V \to \mathrm{Cl}(V, q)$. Mar 4 at 17:11
• @NicholasTodoroff fixed. Mar 4 at 17:16

Yes, it is implicitly understood that we are working with morphisms of unital associative algebras. If we drop the "unital" condition then then things breaks. In the case of $$V = \{0\}$$, the maps $$j$$ and $$k$$ are necessarily the unique maps that take $$0$$ to $$0$$. If we require $$\psi$$ to be unital then $$\mathrm{Cl}(V,q) \cong \mathbb F$$. If we don't require it to be unital then this cannot be because we could have $$\psi(1) = 1$$ or $$\psi(1) = 0$$, so $$\psi$$ is not unique. So instead $$\mathrm{Cl}(V,q) = \{0\}$$.
But recall that we have a nice theorem like $$\dim\mathrm{Cl}(V,q) = 2^{\dim V}$$; the above shows that this theorem fails if $$\psi$$ is not required to be unital.