MHB Left & Right Multiplication Maps on Algebras .... Bresar, Lemma 1.25 .... ....

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Multiplication
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Matej Bresar's book, "Introduction to Noncommutative Algebra" and am currently focussed on Chapter 1: Finite Dimensional Division Algebras ... ...

I need help with the proof of Lemma 1.25 ...

Lemma 1.25 reads as follows:View attachment 6261
View attachment 6262
My questions on the proof of Lemma 1.25 are as follows:Question 1

In the above text from Bresar we read the following:

" ... ... Therefore $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$ ... ... "Can someone please explain exactly why Bresar is concluding that $$[ M(A) \ : \ F ] \gt d^2$$ ... ... ?Question 2

In the above text from Bresar we read the following:

" ... ... Therefore $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$

and so $$M(A) = [ \text{ End}_F (A) \ : \ F ]$$. ... ... "Can someone please explain exactly why $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$ ... ...

... implies that ... $$M(A) = [ \text{ End}_F (A) \ : \ F ]$$ ...
Hope someone can help ...

Peter
===========================================================*** NOTE ***

So that readers of the above post will be able to understand the context and notation of the post ... I am providing Bresar's first two pages on Multiplication Algebras ... ... as follows:View attachment 6263
https://www.physicsforums.com/attachments/6264
 
Last edited:
Physics news on Phys.org
Hi Peter,

A little short on time at the moment, but hopefully I can push you in the right direction.

Peter said:
Question 1

In the above text from Bresar we read the following:

" ... ... Therefore $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$ ... ... "Can someone please explain exactly why Bresar is concluding that $$[ M(A) \ : \ F ] \gt d^2$$ ... ... ?

Because we can conclude from Lemma 1.24 that the operators $L_{u_{i}}R_{u_{j}}$ are linearly independent and there are a total of $d^{2}$ of such operators. Thus the space $M(A)$ that contains all linear combinations of such operators must have at least this dimension (in general if a linear space contains $k$ linearly independent vectors, then its dimension must be greater than or equal to $k$). Note: You wrote a strict inequality in your original post when it should really include equality.

Peter said:
Question 2

In the above text from Bresar we read the following:

" ... ... Therefore $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$

and so $$M(A) = [ \text{ End}_F (A) \ : \ F ]$$. ... ... "Can someone please explain exactly why $$[ M(A) \ : \ F ] \gt d^2 = [ \text{ End}_F (A) \ : \ F ]$$ ... ...

... implies that ... $$M(A) = [ \text{ End}_F (A) \ : \ F ]$$ ...

We know the dimension of $\text{End}_{F}(A)$ over $F$ is $d^{2}$. Since $M(A)$ is a linear subspace of $\text{End}_{F}(A)$, its dimension is at most $d^{2}$. Thus we have $d^{2}\leq M(A)\leq d^{2}$.

I hope that helps. Let me know if anything needs clarification.
 
GJA said:
Hi Peter,

A little short on time at the moment, but hopefully I can push you in the right direction.
Because we can conclude from Lemma 1.24 that the operators $L_{u_{i}}R_{u_{j}}$ are linearly independent and there are a total of $d^{2}$ of such operators. Thus the space $M(A)$ that contains all linear combinations of such operators must have at least this dimension (in general if a linear space contains $k$ linearly independent vectors, then its dimension must be greater than or equal to $k$). Note: You wrote a strict inequality in your original post when it should really include equality.
We know the dimension of $\text{End}_{F}(A)$ over $F$ is $d^{2}$. Since $M(A)$ is a linear subspace of $\text{End}_{F}(A)$, its dimension is at most $d^{2}$. Thus we have $d^{2}\leq M(A)\leq d^{2}$.

I hope that helps. Let me know if anything needs clarification.
Thanks for the help GJA ... really appreciate it ...

Just reflecting on what you have said ...

Peter
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top