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Do infinite dimensional Clifford (and/or Grassmann) algebras exist/makes sense? Do you know good references about them?

rschwieb
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riemannium
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3 Answers3

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Infinite dimensional clifford algebras are the setting of David Hestenes' so-called "universal geometric algebra". Hestenes uses this setting to embed vector manifolds---manifolds whose points are vectors in the UGA and thus admit a lot of niceties in terms of vector operations. For instance, if $\mathbf r(x^1, x^2, \ldots, x^n)$ is a vector function of $n$ parameters that defines a vector manifold, then the tangent vectors are indeed $\partial \mathbf r/\partial x^i$ and so on, and this is well-defined in terms of the manifold being embedded in the UGA.

For more about this application of infinite dimensional clifford algebras, see Hestens and Sobczyk's Clifford Algebra to Geometric Calculus.

Muphrid
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There are infinite dimensional Clifford C*-algebras, which do make sense and are connect to quantum mechanics. In fact the algebra of Canonical Anticommutation Relations can be thought of as a Clifford algebra over a complex vector space, so that in field theory (or any quantum theory with infinitely many degrees of freedom) one requires Clifford algebras over infinite dimensional spaces.

Usually the algebra is defined over an infinite-dimensional space with well-defined geometric properties, namely a Hilbert space. For a real Hilbert space $H$ the Clifford algebra $Cl(H)$ is the (unique) C*-algebra generated by unitaries $u(f)$, $(f\in H)$ subject to $$ u(f)u(g)+u(g)u(f)= \langle f,g \rangle$$ $$ u(f)^2=1, \qquad u(f+ag)=u(f)+au(g).$$ Have a look at

D. Shale, F. Stinespring "States of the Clifford algebra", Ann. Math. 80 (1964), pp 365-381

for more details and possibly other references -Ollie

ViktorStein
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user33750
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Sure they make sense, and nothing really differs in the construction.

I'm not handy with the applications themselves, but I can point you to a few references I did read that hint at the applications.

  • Zou. Ideal structure of Clifford algebras Advances in Applied Clifford Algebras February 2009, Volume 19, Issue 1, pp 147-153

  • Wene. The idempotent structure of an infinite dimensional Clifford algebra Clifford Algebras and their Applications in Mathematical Physics Fundamental Theories of Physics Volume 47, 1992, pp 161-164

  • Lounesto & Wene. Idempotent structure of Clifford algebras Acta Applicandae Mathematica July 1987, Volume 9, Issue 3, pp 165-173

My main interest was in the ring theoretic structure of the algebra itself. It's also worth noting that these are talking about the countable dimensional case only. Countable dimension covers most of the ground for applied mathematics, though :)

rschwieb
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  • Can the Clifford algebra be completed into a Hilbert space, while retaining an everywhere-defined product? (I have found that $\sup_{A,B}\frac{\lVert AB\rVert}{\lVert A\rVert\lVert B\rVert}=\infty$.) – mr_e_man Aug 08 '19 at 06:54
  • @mr_e_man That would probably depend on the inner product you have in mind, but which I am not aware of. – rschwieb Aug 08 '19 at 13:46
  • See https://math.stackexchange.com/questions/3198338/inner-product-structure-on-geometric-algebra – mr_e_man Aug 11 '19 at 20:20
  • @mr_e_man have you checked if the Clifford algebra of the completion is complete with respect to your form? That would be the first obvious thing to check. Maybe there is some categorical property that proves it easily. – rschwieb Aug 11 '19 at 22:23
  • Anyway, I suppose my question should be made into a Question... – mr_e_man Aug 11 '19 at 22:35
  • @mr_e_man Because I don’t understand the inner product you pointed me to, I don’t know whether or not your objection is valid. All I am saying is that it is an obvious thing to check. – rschwieb Aug 11 '19 at 22:55
  • @mr_e_man Hilbert spaces have both Hamel and Schauder bases... I don’t see why an observation about one precludes something bout he other. – rschwieb Aug 11 '19 at 22:58
  • I'd appreciate a comment on my inner product answer, so I can make it more understandable. – mr_e_man Aug 11 '19 at 23:10
  • @mr_e_man like I said, there is no way to conclude as you said that “it doesn’t contain infinite sums” Hilbert space has both hamel and schauder bases. The fact that the space is generated by finite linear combinations of one does not negate the fact that countable linear combinations of the other are also in the space. Perhaps you are just expecting the wrong elements in the Schauder basis. – rschwieb Aug 12 '19 at 02:36
  • @mr_e_man Just because it is constructed in terms of a Hamel basis does not automatically preclude it from having a Schauder basis and being complete normed space. That’s my point! – rschwieb Aug 12 '19 at 03:31
  • Let us continue this discussion in chat. – mr_e_man Aug 12 '19 at 03:37
  • I found my counterexample! Define $I_k=e_{4k-3}e_{4k-2}e_{4k-1}e_{4k}$; for example, $I_1=e_1e_2e_3e_4,;I_2=e_5e_6e_7e_8$. These all commute and square to $1$, and otherwise all possible products of them are orthonormal. Now consider, for some $0<r<1$, $$A=\sum_{k=0}^\infty r^k\prod_{j=1}^{k}(1+I_j)=\frac{1}{1-r}+\sum_{k=1}^\infty \frac{r^k}{1-r}I_k\prod_{j=1}^{k-1}(1+I_j)$$ The norm $\lVert A\rVert^2$ converges if $r<2^{-1/2}\approx0.71$, but $\lVert AA\rVert^2$ diverges if $r\geq2^{-3/4}\approx0.59$, so the completed Clifford algebra is not closed under multiplication. – mr_e_man Aug 16 '19 at 05:21
  • Likewise, noting that $(1+I_k)\wedge(1+I_k)=(1+2I_k)$ which has norm-squared $1^2+2^2=5$, we can find that $\lVert A\wedge A\rVert^2$ diverges if $r\geq5^{-1/4}\approx0.68$, so the completed Clifford algebra is also not closed under the wedge product. ...If we complete each grade individually, but not the whole algebra, I suspect that it would be closed under both products. – mr_e_man Aug 19 '19 at 00:36