Inductive proof for multiplicative property of sdet

In summary, the conversation discusses the proof of the multiplicative property of super determinant in Roger's book on supermanifolds. The proof involves inverting matrices that can be represented as formal power series, resulting in complex expressions. One person is struggling with the technical mess and asks if anyone has tried this method before. They provide a summary of their struggle, including the shorthand notation used in the computation. They have tried to show the equality order by order but have found it to be too involved.
  • #1
Korybut
64
3
TL;DR Summary
Is it doable?
Hello!

Reading Roger's book on supermanifolds one can find sketch of the proof for multiplicative property of super determinant. Which looks as follows
Sdet.jpg

All the words sounds reasonable however when it comes to the direct computation it turns out to be technical mess and I am about to give up. I mean all the matrices that need to be inverted are invertible and can be represented as formal power series so everything is well-defined nonetheless resulting expressions are of insane technical complexity. Have someone tried this particular way of proving this property?
 
Physics news on Phys.org
  • #2
To add some detail of my struggle. I represent my super matrices according to suggested in the proof way
##\mathcal{M}=\left(\begin{matrix}M_{00}+\beta_L A_{00} & M_{01}+\beta_L A_{01} \\
M_{10}+\beta_L A_{10} & M_{11}+\beta_L A_{11}\end{matrix}\right)##

##\mathcal{N}=\left(\begin{matrix}N_{00}+\beta_L B_{00} & N_{01}+\beta_L B_{01} \\
N_{10}+\beta_L B_{10} & N_{11}+\beta_L B_{11}\end{matrix}\right)##
Since it is clear that result will be only linear in ##A## or ##B## I will provide only the part that comes with ##A_{01}## (nothing particular special about ##A_{01}##. I just want to see what comes here). After straight-forward computation one has
##sdet\mathcal{M}\, sdet \mathcal{N}=sdet M\, sdet N\, \left(1-\beta_L A_{01}M_{11}^{-1} M_{10} \left(M_{00}-M_{01} M_{11}^{-1} M_{10}\right)^{-1}+\ldots \right)##
I know that one should actually "take the trace" but I will omit the indices for simplicity. Appearing here matrix ##M_{00}-M_{01} M_{11}^{-1} M_{10}## is invertible since ##M_{01} M_{11}^{-1} M_{10}## has no body. Its inverse looks as follows
##\left(M_{00}-M_{01} M_{11}^{-1} M_{10}\right)^{-1}=\sum_{p=0}^\infty \left(M_{00}^{-1} M_{01} M_{11}^{-1} M_{10}\right)^p M_{00}^{-1}##
Series has only finite amount of terms so one doesn't need to worry about convergence properties.

On the other hand one can compute ##sdet \mathcal{M} \mathcal{N}## and again examine what comes with ##A_{01}##. Since resulting formula is very involved I introduce following shorthand notation
##\mathcal{X}=M_{00} N_{00}+M_{01}N_{10}-\left(M_{00}N_{01}+M_{01}N_{11}\right) \left(M_{10}N_{01}+M_{11}N_{11}\right)^{-1}\left(M_{10} N_{00}+M_{11}N_{10}\right)##
This one of the parts that comes computing ##sdet MN##.

##\mathcal{Y}=N_{10}-N_11 \left(M_{10}N_{01}+M_{11}N_{11}\right)^{-1}\left(M_{10}N_{00}+M_{11}N_{10}\right)##.
Once again all the matrices that need to inverted are invertible as a formal series.

Using these notation result for computation ##sdet \mathcal{M}\mathcal{N}## looks as follows
##sdet \mathcal{M}\mathcal{N}=sdet MN\, \left(1+\beta_L A_{01} \mathcal{Y} \mathcal{X}^{-1}+\ldots \right)##

I've tried to show equality order by order in ##N_{01}##. It is not that hard to show that in zeroth order (i.e. ##N_{01}=0##) equality holds. However when it comes to first order it turns out to be too involved.
 

Related to Inductive proof for multiplicative property of sdet

1. What is an inductive proof?

An inductive proof is a mathematical method used to prove a statement or theorem for all possible values of a variable. It involves showing that the statement is true for a specific value, and then using that to prove that it is also true for the next value, and so on.

2. What is the multiplicative property of sdet?

The multiplicative property of sdet (short for symmetric determinant) states that the determinant of a matrix A multiplied by the determinant of its transpose is equal to the determinant of the product of A and its transpose. In other words, det(AB) = det(A) * det(B).

3. How is the multiplicative property of sdet proved using induction?

The proof involves showing that the property holds for a 1x1 matrix, and then using that to show that it also holds for a 2x2 matrix. This process is repeated for larger matrices, with the assumption that the property holds for matrices of a certain size being used to prove it for the next size. This is continued until it can be shown that the property holds for all possible sizes of matrices.

4. Why is the inductive proof important for the multiplicative property of sdet?

The inductive proof provides a rigorous and systematic way of proving the multiplicative property of sdet for all possible sizes of matrices. It also helps to establish the validity and reliability of the property, making it an essential tool for mathematicians and scientists working with matrices.

5. Are there any limitations to the inductive proof for the multiplicative property of sdet?

While the inductive proof is a powerful tool for proving the multiplicative property of sdet, it does have its limitations. It can only be used for finite matrices and may not be applicable for infinite matrices. Additionally, it may not be the most efficient method for proving the property, as there may be alternative proofs that are more concise or elegant.

Similar threads

  • Programming and Computer Science
Replies
10
Views
943
Replies
3
Views
354
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • STEM Educators and Teaching
Replies
3
Views
2K
Replies
2
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
2K
Replies
6
Views
4K
Replies
7
Views
1K
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
2K
Back
Top