Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is a follow-up question to When does a symmetric algebra over a field of characteristic 0 fail to be semisimple?

Let $H$ be a symmetric algebra over $\mathbb{R}$ with symmetrizing trace $\tau:H\to \mathbb{R}$ such that the corresponding bilinear form $H \times H \to \mathbb{R}, (h,h') \mapsto \tau(hh')$ is positive definite. Assume also that $H$ is split over $\mathbb{R}$. Then we can choose an orthonormal basis $\mathcal{B}$ for the bilinear form. Thus for each simple $H$-module, the Schur element $c_V$ is given by $$ c_V = \frac{1}{\dim_\mathbb{R} V} \sum_{b \in \mathcal{B}}\chi_V(b)^2 \geq 0. $$ Moreover, $c_V = 0$ implies that $\chi_V(h) = 0$ for all $h \in H$, but this contradicts the assumption $V \neq 0$.
Hence all the Schur elements are nonzero, so $H$ is semisimple.

Does anyone know a reference for this result?

Does anyone know of generalizations of this result? Specifically, is there a more general setting in which positive definiteness can be defined and can this still be used to conclude that $H$ is semisimple? For instance, suppose $H$ is an algebra over $\mathbb{R}(q)$, for an indeterminate $q$. Does the result hold without the assumption that $H$ is split?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

There is a lemma due to Dieudonné that is similar and may be the generalization you are seeking. (My reference is Nathan Jacobson's book "Structure and Representations of Jordan algebras", in Chapter VI on pages 239, 240.) We replace the real numbers with a field $F$---this can be any field.

Let $A$ be a finite-dimensional and possibly nonassociative $F$-algebra. (This means that $A$ is a vector space and there is a bilinear map $A \times A \to F$ that we view as the product on $A$. We don't make any other assumptions about the product.)

Dieudonné says:

If there is a nondegenerate and associative bilinear form $b$ on $A$, and $A$ has no nonzero ideals $I$ such that $I^2 = 0$, then $A$ is a direct sum of simple ideals.

To fill in a couple of definitions: The form $b$ is nondegenerate (a generalization of "positive definite") if for every nonzero $x \in A$ there is a $y \in A$ so that $b(x, y)$ is not zero. It is associative if $b(xy, z) = b(x, yz)$ for all $x,y,z \in A$. This $b$ plays the role of your $\tau$.

Added

If you assume that the multiplication on $A$ is associative, then each element comes with a "generic minimal polynomial" which mimics the properties of the characteristic polynomial in the case of $n$-by-$n$ matrices. The linear coefficient gives an analogue of the trace $tr: A \to F$ which lets us define a symmetric bilinear form $t(a, a') := tr(aa')$. This form is obviously "associative" in the above sense (because $A$ is associative). Furthermore, if $I$ is an ideal of $A$ such that $I^2 = 0$, then every $i \in I$ and $a \in A$ satisfy $t(i, a) = tr(ia)$, which is 0 because $ia$ is nilpotent. Therefore, as Mariano points out, one finds (without Dieudonné):

If the bilinear form $t$ is non-degenerate, then $A$ is a direct sum of simple ideals.

This provides the condition for semisimplicity in terms of a bilinear form that you were asking for.

share|improve this answer
    
This may be helpful, thank you. To show semisimplicity of $A$, we would still need to show that $A$ has no ideals $I$ such that $I^2=0$, but this seems easier than showing that the Schur elements are nonzero. –  Jonah Blasiak Jun 28 '11 at 2:51
3  
For an finite dimensional associative algebra over a field, knowing that there are no ideals $I$ with $I^2=0$ is enough to show that the algebra is semisimple, regardless of any bilinear form. –  Mariano Suárez-Alvarez Jun 28 '11 at 5:11
    
And your addendum is wrong. There are symmetric algebras that are not semisimple. See for example the other thread mathoverflow.net/questions/68883 for a construction –  Johannes Hahn Jun 30 '11 at 10:44
    
@Johannes: My addendum is correct. I specified that you must construct the bilinear form from the generic trace on the algebra; this is a naturally determined linear map $A \to F$, which in the case of the dual numbers $A = F[x]/(x^2)$ is given by $tr(a+bx) = 2a$ for $a, b \in F$, so the bilinear form $t$ is degenerate. The nondegenerate bilinear form appearing in Jack Schmidt's example that you point to is a different one. –  Skip Jul 1 '11 at 4:05

Though not an answer to the request for references, I'd like to add the fact that these algebras are trivial:

If $A$ is a split, symmetric $\mathbb{R}$-algebra with positive definite trace form (in the above mentioned sense), it is semisimple as proven above. It is therefore a product of matrixrings over $\mathbb{R}$ and the traceform is a linear combination with positive coefficients of the traces of these matrixrings. But the trace on $\mathbb{R}^{d\times d}$ is not positive definite if $d>1$ because there are nonzero matrizes with square zero. Therefore we get an algebra isomorphism $A \cong \mathbb{R} \times \ldots \times \mathbb{R}$.

If there is indeed a generalization of this result to other fields than $\mathbb{R}$, this proof would probably carry over.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.