Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I extract a problem from the book I am reading:

Let $R$ be a field, $A$ be a semisimple split $R$-algebra (associative with $1$). Let $A = \oplus_{n=1}^t A_n$ be a decomposition of $A$ into simple algebras. $(,): A \times A \rightarrow R$ is a non-degenerate symmetric $R$-bilinear form. Then $(A_i, A_j) =0$ for $i \neq j$.

Is this statement true? If it is, why? If not, how can I find a conunterexample? In the first place, what is the definition for a split algebra? I think of split extension of fields, but there seems to be no realtion.

Thanks very much.

share|improve this question
    
Shouldn't you tell us what the definition of split algebra in the book that you are using but don't name is? –  Rasmus Aug 8 '12 at 13:55
    
@Rasmus: Thank you very much for the comment. In fact, this book(pages 74,75 of arxiv.org/pdf/math/0208154v1.pdf) doest not give the definition of a split algebra. Maybe Mr. Lusztig assumes that all the readers know about it. –  ShinyaSakai Aug 9 '12 at 9:49
add comment

1 Answer

up vote 2 down vote accepted

An $R$-algebra $A$ being split means that ${\rm End}_A M = R$ for any simple $A$-module $M$, to the best of my knowledge. In any case, statement is false though. If any of the $A_i$ are not distinct, then we can twist the bilinear form a little, while preserving the desired properties, and end up with a counterexample. An explanation follows, but it would be worth considering it yourself first with say, the case $A=S\oplus S$ for some simple $R$-algebra $S$.



Take for example $A=Re_1\oplus Re_2$ to be the semisimple $R$-algebra, where $e_1,e_2$ are orthogonal idempotents.

This is split because the simple $A$-modules are just $e_1A$ and $e_2A$ and so it is apparent that the endomorphisms of these are just multiples of the identity.

Now define a bilinear form on $A$ by $(x,y)=x_1y_2+x_2y_1$ where $x=x_1e_1+x_2e_2$ and $y=y_1e_1+y_2e_2$ with $x_1,x_2,y_1,y_2\in R$. Then this bilinear form is symmetric and nondegenerate since $(x,e_2)=x_1$ and $(x,e_1)=x_2$. By construction, the simple subalgebras are isotropic (i.e. $(e_i,e_i)=0$) but the inner product between them is nonzero (i.e. $(e_1,e_2)=1$).

By the way, a more concrete way to see all of this is that this inner product is given by $$\begin{bmatrix}x_1 & x_2\end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}y_1 \\ y_2\end{bmatrix},$$ and nondegeneracy and symmetry follow from the fact the matrixis invertible and symmetric.

share|improve this answer
    
Thank you very much. I think I must have missed some conditions. In fact I am reading (arxiv.org/pdf/math/0208154v1.pdf), and this problem comes from the begining of 19.1 (on page 74) and the second paragraph of page 75. I think $A_i,A_j$ for $i \neq j$ may not be distinct (that is, they may be isomorphic), but the "trace form" makes the difference. –  ShinyaSakai Aug 9 '12 at 9:47
    
Indeed! The trace form ensures that the product of elements in distinct factors are zero, hence their inner product is zero. –  user1306 Aug 9 '12 at 13:12
    
Thank you. But this trace form is not defined classically. It is defined by the bilinear form. (A trace form is an $R$-linear map $\tau: A \rightarrow R$ such that $(a,a') = \tau(aa') = \tau(a'a)$ is a non-degenerate (symmetric) $R$-bilinear form $(,): A \times A \rightarrow R$.) Then, why is it obvious that $(A_i,A_j) = 0$ for $i \neq j$? Thanks again. –  ShinyaSakai Aug 9 '12 at 15:46
    
Right, but that's the point: if $A=\bigoplus A_i$ and we have $a\in A_j$ and $b\in A_k$ for $j\neq k$, then $(a,b)=\tau(ab)=\tau(0)=0$ ($ab=0$ since they lie in different summands and $\tau(0)=0$ by $R$-linearity). The important part is that it is defined by first taking the product, then some $R$-linear map. –  user1306 Aug 10 '12 at 0:02
add comment

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.