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Suppose given a locally convex Hausdorff topological vector space $V$ over $\mathbb R$ and a continuous, symmetric, bilinear map $q:V\otimes V\to \mathbb R$, where the tensor product is the completed projective tensor product (in the case I am interested in, $V$ is the space of smooth sections of a smooth vector bundle on a compact manifold.) One can define a Clifford algebra $Cl(V,q)=\oplus_k V^{\otimes k}/I$, where $I$ is the topological ideal generated by elements of the form $x\otimes y+y\otimes x -2q(x,y)$. My first question is: how does one guarantee that $I$ is closed and therefore that the quotient is Hausdorff? Second, assuming that $CL(V,q)$ is Hausdorff, does $Cl(V,q)$ satisfy the universal property that, given any algebra object $A$ in the category of locally convex topological vector spaces equipped with the completed projective tensor product, and any continuous map $\varphi: V\to A$ satisfying $\phi(x)\phi(y)+\phi(y)\phi(x)=2q(x,y)$, there exists a unique continuous map of algebras $Cl(V,q)\to A$?

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    $\begingroup$ I reckon simply take closure of $I$ if it is not closed to begin with. If $A$ is Hausdorff, and $I$ is sent to zero by the unique extension of a (linear) continuous $\varphi$ to $Cl$, then the closure of $I$ will go to zero too $\endgroup$ Aug 23, 2018 at 18:21
  • $\begingroup$ Hmm, good point. Is the closure of $I$ still an ideal? $\endgroup$ Aug 23, 2018 at 20:22
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    $\begingroup$ This is fairly standard. E. g. If $ab\notin$ closure, i. e. there is an open $U\ni ab$ with (closure) $\cap U=\varnothing$; then $b\notin$ closure since $I\cap \left(a^*U\right):=I\cap\{x\mid ax\in U\}=\varnothing$, with $b\in a^*U$, and the latter set is open since (left) multiplication by $a$ is continuous. $\endgroup$ Aug 24, 2018 at 4:38
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    $\begingroup$ Alternatively just take intersection of closed ideals containing $I$ -- even without knowing that it coincides with closure you may use it in your situation instead of the closure. $\endgroup$ Aug 24, 2018 at 4:39
  • $\begingroup$ Ah, that makes a lot of sense. Thank you! $\endgroup$ Aug 24, 2018 at 12:36

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