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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.

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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 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
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.

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Thank you very much. I think I must have missed some conditions. In fact I am reading (, 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

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