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Suppose you have a finite dimensional real Hilbert space $V$ and you form the tensor algebra $$T(V) = \mathbb{R} \oplus V \oplus (V\otimes V) \oplus (V\otimes V \otimes V) \oplus \cdots$$ where the multiplication is given by the tensor product. Now imagine you have a finite set of vectors $S$ which you use to form an ideal $I(S)$. From this, we can get the quotient algebra $Q = T(V)/I(S)$ (see this Wikipedia article).

I am interested in comparing elements of $Q$, ideally using an inner product, if it is possible to define one, or else using a distance, if it is possible to define a norm.

The approach I've taken so far is, given an element $x$ of $T(V)$ is to try to project it onto the orthogonal complement of $I(S)$ (call this $P(x)$) in the assumption that there will be a unique element of the equivalence class of $x$ that lives in this orthogonal complement, and that we can then use the inner product of $T(V)$. However I think this assumption is false. Consider the case where $S =$ {$a\otimes a - a$} for a one dimensional vector space $V$ with basis vector $a$. Then $I(S)$ is the vector space with basis $a - a\otimes a$, $a\otimes a - a\otimes a\otimes a$, $a\otimes a\otimes a - a\otimes a\otimes a\otimes a$, $\ldots$.

Intuitively, I would expect $Q$ in this case to be a two dimensional vector space with basis $1, [a]$. Is this correct? Projecting onto the orthogonal complement seems to give zero however.

Is there a standard way to define a norm or an inner product on a quotient algebra such as this? How can it be computed?

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Is it obvious that T(V) itself inherits a norm or an inner product from V? –  Qiaochu Yuan Jan 22 '10 at 17:26
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Oh, I see. Are you talking about the inner product where tensors of different degree are orthogonal and (a \otimes b, c \otimes d) = ((a, c) * (b, d), and so forth? –  Qiaochu Yuan Jan 22 '10 at 18:21
    
Yes - that is the inner product I had intended. –  Daoud Jan 22 '10 at 22:05

3 Answers 3

up vote 3 down vote accepted

Intuitively, I would expect Q in this case to be a two dimensional vector space with basis 1,[a]. Is this correct? Projecting onto the orthogonal complement seems to give zero however.

The first part is correct. This may be easiest to see by considering the isomorphism, in the present case where $V$ is one dimensional, of $T(V)$ with $\mathbb{R}[x]$. Then $Q$ is $\mathbb{R}[x]/(x^2-x)$, which is 2-dimensional by the division algorithm.

The orthogonal complement of $I(S)$ in this case is not zero, but it is one dimensional and your concern is valid. Thinking again in $\mathbb{R}[x]$ for convenience, the condition that $\sum_k a_k x^k$ is orthogonal to $x^{m+2} - x^{m+1}$ for all $m\geq0$ says that $a_{m+2}=a_{m+1}$ for all $m\geq0$; thus, $a_k=0$ for $k\geq1$. Note that the conclusion is the same if we first complete to a Hilbert space by taking power series with square summable coefficients, so that unfortunately Andrew Stacey's desperate hope appears to be shattered. I have been assuming in this paragraph that $a$ (or $x$) has norm 1; it makes no difference for the conclusion in the non-completed case, but it simplifies the relation. (In other cases when $\dim(V)>1$ the Hilbert space quotient can be far too large for the vector space isomorphism to be possible, having dimension $2^{\aleph_0}$, so completion is probably not the way to go.)

This shows that there is no hope in general of using the standard inner product to define a norm on $T(V)/I(S)$. However, the problem in the above case arose because the ideal was not homogeneous; this forces relations on coefficients of different degree for elements of the orthogonal complement, thereby forcing too many coefficients to be zero. For this construction to work I recommend considering the case where $S$ is a set of homogeneous elements of $T(V)$ (i.e., each element of $S$ lies in one of the tensor powers $V^{\otimes k}$). I don't know what else to tell you in the nonhomogeneous case.

Gratuitous Add-on

The Hilbert space completion of the free algebra $T(V)$ is the full (or free) Fock space $F(V)$. The latter is not an algebra, but a Banach algebra completion of $T(V)$ can be obtained by first representing $T(V)$ on $F(V)$ by choosing an orthonormal basis $v_1,\ldots,v_n$ for $V$ and sending $v_j\in T(V)$ to the "creation operator" $S_j$ on $F(V)$ defined by $S_j(w)=v_j\otimes w$, and then closing the image in your favorite topology. These algebras were first studied by G. Popescu circa 1991 (but over $\mathbb{C}$ rather than $\mathbb{R}$ and with either norm or ultraweak closure), and a similar construction was earlier used by D. Evans to study the Cuntz algebra $\mathcal{O}_n$ in the 1980 paper "On $\mathcal{O}_n$".

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Nice answer. Just a remark in passing that the Banach algebra defined in this construction is not the universal tensor algebra of $V$; there one would take the $l^1$-direct sum of projective tensor powers of $V$. That construction predates Popescu's, but I don't know that anyone ever managed to do anything with it –  Yemon Choi Jan 23 '10 at 19:15
    
Thanks. That is very interesting, and I was unaware of that construction. (My gratuitousness reveals both my operator algebraic bias and my ignorance, which is good because I learned something new.) To compare the different constructions when dim(V)=1: Popescu's are isometrically isomorphic to the disk algebra and H^∞, whereas the construction you refer to leads to $l^1$(ℕ). Will you please point me to a reference for these $l^1$ tensor algebras? –  Jonas Meyer Jan 23 '10 at 22:10
    
Thanks for answering this in such detail. Unfortunately, the use of non-homogeneous ideals is central to what I want to do. The Fock space idea had occurred to me also - I'll write about this in a separate answer. –  Daoud Jan 23 '10 at 22:14
    
I feared that would be the case, and I hope that someone knows a construction that will help you. I look forward to your answer. –  Jonas Meyer Jan 23 '10 at 22:23
    
@Jonas: I should really have said "the construction almost certainly predates Popescu's" since I can't recall a precise reference off the top of my head (last time I thought about this properly was en passant during my PhD). The l^1-flavoured construction is, if you like categorical stuff, the left adjoint of the forgetful functor from unital BAs to B. spaces –  Yemon Choi Jan 24 '10 at 3:27

I'm not sure if this is what you want, but what I would do is to first complete $T(V)$ with respect to (some) inner product, say the one that Qiaochu says in the comments where the different pieces are orthogonal and have the obvious determinant inner products on them. Then I could merrily use Hilbert space theory with aplomb. In particular, the quotient $\overline{T(V)}/\overline{I(S)}$ is isomorphic to the orthogonal complement of $I(S)$ in $\overline{T(V)}$. Then I would hope - possibly desperately - that $S$ was such that the algebraic quotient, $T(V)/I(S)$, was isomorphic to the completed quotient $\overline{T(V)}/\overline{I(S)}$.

(I haven't worked through this, so I don't know if it works for your example, but the finite dimensionality of the quotient gives me hope that it does!)

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Extending Jonas' Gratuitous Add-on, the Fock space idea seems promising. Since it is a Hilbert space, the space of operators will form a C* algebra, and quotient algbras of C* algebras are again C* (see e.g. Toeplitz matrices, asymptotic linear algebra, and functional analysis by Albrecht Böttcher and Sergei M. Grudsky, 2000) so the norm is well defined.

Of course the ideal in this case is an entirely different beast since we have to include multiplication by annihilation operators as well. If we write $|a\rangle$ and $\langle a|$ for the creation and annihilation operators respectively corresponding to $a \in V$ then the ideal generated by $|a\rangle|a\rangle - |a\rangle$ includes for example $|a\rangle - 1$ and $1 - \langle a|$ where I have multiplied through by the annihilation operator $\langle a|$.

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I don't see why you would necessarily want to throw in the annihilation operators; couldn't you work in non-self-adjoint algebras rather than C* algebras? In either case, to have a Banach algebra norm on the quotient you will need to close your tensor algebra and ideal, and I do not see how this will relate directly to the orginal construction. –  Jonas Meyer Jan 23 '10 at 23:06
    
I'm afraid I didn't understand this part of your answer (I must confess I'm not a mathematician). Perhaps you could point me to a reference that would help me understand why closing causes such a problem. –  Daoud Jan 23 '10 at 23:10
    
I don't know a good reference for discussing this phenomenon in general, but the problem is this: Your problem is about a purely algebraic object, Q. If you translate the problem to a functional analysis problem (and it is not even clear to me what the best way to do that is), then it is a different problem that may not help you in your original problem. If you could find a way to embed Q faithfully in a quotient Banach algebra then you would get a norm, but this won't always be possible because in the Banach algebra quotient you mod out by a bigger ideal (that is the problem of closing). –  Jonas Meyer Jan 23 '10 at 23:37
    
There is a reason I called that part of my answer "gratuitous". The construction came to mind and I couldn't resist mentioning it, but I had no reason to think it would have bearing on solving your problem. But perhaps the translated problem(s) would interest you too? –  Jonas Meyer Jan 23 '10 at 23:44
    
They could well do. The application is a very practical one in computing. The specific properties of the vector space are up for grabs - the main requirement is simply that we can compute the norm on the quotient algebra. If you're interested, perhaps we could discuss this via email, as I have a feeling we may have to go off-topic (I found your home page). Thanks again for your help. –  Daoud Jan 24 '10 at 1:15

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