Dan Popescu asked me the following question, and since I'm not an expert I'm throwing his question on MO.

Suppose that $A$ is a finite-dimensional vector space over an ordered field $k$ with $\operatorname{char}(k) = 0$, equipped with:

  • a commutative associative $k$-linear multiplication $\circ\\,$;
  • a positive-definite inner product $\langle \cdot , \cdot \rangle \\,$.

For each $a \in A$, let $L_a \colon A \to A \colon x \mapsto a \circ x$. This is a linear operator on $A$; consider the adjoint operator w.r.t. the given inner product:

$$\langle L_a x, y \rangle = \langle x, L_a^* y \rangle$$ for all $x,y \in A$. Now write $y \star a := L_a^* y$.

Popescu's question is the following:

Is there a name for this operation $\star$? Has it been studied in the literature? Is there anything known about algebraic properties or identities involving $\star$ (possibly also involving the other data $\circ$ and $\langle \cdot , \cdot \rangle$)?


3 Answers 3


In case $k=\mathbb{C}$, what you're describing is a finite-dimensional H*-algebra. More generally, these are Banach algebras, whose carrier space is a Hilbert space, satisfying the adjoint property you mention.

It is natural to make a nondegeneracy assumption: $A$ is called proper when $\forall a \in A\\,.\\, a \circ A = 0 \Rightarrow a = 0$. This turns out to be equivalent to the adjoint $L_a^\ast$ being unique, or, in other words, $\ast$ being an involution. Every H*-algebra is a direct sum of a proper one and an algebra in which $a \circ b=0$ for all $a$ and $b$.

There is a neat structure theorem by Warren Ambrose (see http://www.jstor.org/stable/1990182), showing that proper H*-algebras are always direct sums of full matrix algebras. In particular, commutative H*-algebras are direct sums of 1-dimensional algebras, and hence correspond precisely to orthogonal basis of their carrier space!

I don't know about other fields $k$, but Ambrose's proof basically comes down to carefully analyzing idempotents, which is feasible to repeat for other fields $k$.

  • 1
    $\begingroup$ In $H^*$-algebras, the adjoint of a multiplication operator $L_a$ is always itself a multiplication operator. Is this automatic in the setting that I described, or is this an additional assumption? $\endgroup$ Aug 25, 2011 at 14:47
  • $\begingroup$ (As pointed out by Dima Shlyakhtenko in his answer ---which is in fact a response to this comment--- this is indeed an additional assumption.) $\endgroup$ Aug 26, 2011 at 9:02
  • $\begingroup$ I'll accept this answer; it's not exactly what I had in mind, but it's definitely very useful --- thanks! $\endgroup$ Aug 29, 2011 at 7:00

(This is too long for a comment, my apologies for posting this as an answer).

I don't think it is in general true that $L_a^*$ is a multiplication operator. For example, let $S$ be an invertible $n\times n$ matrix, and consider $$A=\{ SDS^{-1} : D \textrm{ diagonal}\}$$ as an abelian subalgebra of $n\times n$ matrices $M_{n\times n}$. Put an inner product on $M_{n\times n}$ by $\langle X,Y\rangle = Tr(XY^*)$, and endow $A$ with its restriction. Finally, let $P : M_{n\times n}\to A$ be the orthogonal projection onto $A$.

For a matrix $X\in M_{n\times n}$ denote by $\mathscr{L}_X$ the operator $$\mathscr{L}_X : M_{n\times n} \ni Y \mapsto XY \in M_{n\times n}.$$ It is not hard to see that $\mathscr{L}_X^* = \mathscr{L}_{X^*}$.

Then for $a = S D S^{-1} \in A$ your $L_a$ is given by $L_a = P \mathscr{L}_{SDS^{-1}} P $ and so its adjoint (as a linear map from $A$ to $A$) is given by $$P \mathscr{L}_{ (S^{-1})^* D S^*} P$$ and its value on some $b = SD' S^{-1}$ is given by $P( (S^{-1})^*DS^* S D'S^{-1})$ so that $L_a^*$ is no longer a multiplication operator.

It looks like, at least for abelian subalgebras of $M_{n\times n}(\mathbb{C})$ with the given inner product,the necessary and sufficient condition for $L_a^*$ to belong to $L_A$ is that $A$ be closed under the adjoint operation, i.e. be a von Neumann algebra.

  • $\begingroup$ Thanks for this explicit counterexample; it was indeed my intuition that this condition would not be automatically fulfilled. That brings me back to the original question... $\endgroup$ Aug 26, 2011 at 7:58
  • $\begingroup$ Nice example! The last paragraph confuses me, though. All finite-dimensional von Neumann algebras are direct sums of full matrix algebras, and hence must be H*-algebras. So that cannot be the characterizing condition for each $L_a^*$ to be a multiplication operator again. But the norm induced by the trace inner product does not satisfy the C*-condition $\|X\|^2=\|X^*X\|$, so at least there is no contradiction. $\endgroup$ Aug 26, 2011 at 10:14
  • $\begingroup$ @Chris: what I meant was that you are embedding $A$ into linear operators $End(A)$ on the vector space $A$ and that you have (by the choice of your inner product on $A$) chosen a $*$-operation on $End(A)$; together with the operator norm $\Vert T\Vert = \sup_{\langle xi,xi\rangle=1}\langle A\xi ,A\xi\rangle$ and this $*$-operation, $End(V)$ is a von Neumann algebra. To require that $L_a^* \in L_A$ you end up requiring that the image of $A$ (under $a\mapsto L_a$) is closed under the $*$-operation on $End(V)$, i.e. be a von Neumann subalgebra of $End(V)$. $\endgroup$ Aug 27, 2011 at 2:43

This operation has been extensively studied for the needs of (weighted) automata theory.

In the case of the free algebra (this is a non-commutative version of the question of the MO) Here
pp 8 (end) and 9 (18 and 19 of the .pdf) and in the case of the shuffle algebra (which is commutative).

Note that, if the adjoints $L_a^{\star}$ are computed within the algebraic dual, the linear forms which have a finite dimensional orbit by the collection $(L_a^{\star})_{a\in A}$ is exactly the Sweedler's dual of the algebra $A$ (it is therefore a coalgebra). See, e.g. Sweedler's duals and Schützenberger's calculus

If you are interested do not hesitate to interact.


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