Norm on quotient algebra of a tensor algebra 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?
 A: 
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 $\mathcal{O}_n$".
A: 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!)
A: 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 $\langle a|$.
