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 am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1 .

I have a question on page 9, line -7 (see Page 9 here). It is said that $$f_X(x_2) = x_2e_{21} = x_2e_{21}e_{11}.$$ It seems that $e_{11}$ has not been defined and $e_{21}$ is a $2 \times 2$ matrix. How can $e_{21}$ act on $x_2$? Thank you very much.

Edit: On -9, how to show that $x_1a_{11} + x_2a_{21}$ is an element in $X_1$? It seems that $x_2a_{21}=f_X(x_2)a_{21}$. But this cannot be true since the left hand side is in $X_2$ and the right hand side is in $X_1$.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

My guesses are that:

1) $e_{11}=e_1$, i.e. a typo. I would have used the notation $e_{11}$ and $e_{22}$ in place of $e_1$ and $e_2$ to make it easier to remember that they really are 2x2 matrices. OTOH $e_1$ and $e_2$ are standard notations for idempotents, so I understand the authors choice as well. It is possible that the author has changed the notation at some point, and no one noticed to make the change here.

2) Here there is another typical abuse of notation. The symbol $x_2$ stands for a vector in the space $X_2$ (similarly $x_1\in X_1$). When we agree to talk about the space $X$ as a direct sum $X_1\oplus X_2$ of its two subspaces we need to pay the price of making certain identifications. A vector $x_2\in X_2$ can be viewed as an element of $X$. because $X_2$ is a subspace of $X$. But when we write $X=X_1\oplus X_2$, and accordingly use the vector notation $(x_1,x_2)$ for its elements, we must identify $x_2$ with $(0,x_2)$. Similary we must identify all elements $x_1\in X_1$ with vector $(x_1,0)$. Early on in algebra studies a distinction is made between inner and outer direct sum. When a certain maturity level is reached, the distinction is blurred. The author may have jumped to this point too soon, or may be not. I cannot tell for sure. Anyway, once you think of $x_2$ as the vector $(0,x_2)$ whenever necessary, your problem should disappear.

share|improve this answer
    
thank you very much. –  LJR Mar 6 '13 at 9:26
    
but on line -9, $(x_1a_{11}+x_2a_{21}, x_2a_{22}) = (x_1a_{11}+f_X(x_2)a_{21}, x_2a_{22})$. Here $x_1a_{11}+x_2a_{21}$ should be an element in $X_1$. But $x_2a_{21}$ is in $X_2$. –  LJR Mar 6 '13 at 10:55
1  
I understand now. The $x_2$ in $f_X(x_2)$ is indeed $(0, x_2)$ but the $x_2$ in $x_2a_{21}$ is an element in $K$. –  LJR Mar 6 '13 at 11:00
    
The $x_2$ in $f_X(x_2)$ is indeed $(0, x_2)$. But how to show that $x_1a_{11} + x_2a_{21}$ is an element in $X_1$? It seems that $x_2a_{21}=f_X(x_2)a_{21}$. But this cannot be true since the left hand side is in $X_2$ and the right hand side is in $X_1$. –  LJR Mar 6 '13 at 11:07
    
$X_2$ is a subspace, it is not an $A$-submodule. Right multiplication by $a_{21}e_{21}$ does take $X_2$ into $X_1$: $$(0,x_2)\pmatrix{0&0\cr a_{21}&0\cr}=(x_2a_{21},0).$$ –  Jyrki Lahtonen Mar 6 '13 at 11:27

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.