Is this a good proof of Schur's Lemma?

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In summary, the conversation discusses the application of Schur's lemma, specifically the version involving irreducible representations, to a particular equation involving the operators A and D(g). The speaker shares their own result and notes that there are two possibilities, but only one is true in general for any group element. The conversation also mentions the need for additional information and clarification on the variables and assumptions involved in order to fully understand the application of Schur's lemma in this context.
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kent davidge
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There are plenty of proofs of Schur's lemma on the internet, but I find them hard to follow. Then I came up with my own result, but I'm not sure if it's good enough.

Consider ##A v = \kappa v## and ##A v=\kappa v ##. Operating with ##D(g)## the equation then becomes ##D(g)A v = \kappa D(g) v## or ##D(g)(A - \kappa I) v = 0##. But ##A## and ##D(g)## commute, so ##(A - \kappa I)D(g) v = 0##.

Then there would be two possibilites. The first is that the parenthetic quantity vanishes and the second ##D(g)## anihilates ##v##. But this is not true in general for any ##g##. Then the first case would hold, i.e., ##A = \kappa I##.
 
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Oops I should have said that ##D(g)## is irreducible representation. I guess it's completely possible for a reducible representation to map ##v## to the zero vector for any group element ##g##, that is anihilate it, that is let the space consisting of the zero vector invariant.
 
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kent davidge said:
Oops I should have said that ##D(g)## is irreducible representation
You should also have stated which version of Schur's Lemma you are talking about, what ##A,D,v,g,\kappa## are, why ##A## commutes with ##D(g)## and why you can conclude that ##A-\kappa I = 0## only because it vanishes for a certain element ##w:=D(g)(v)## which you have specified by ##Av=\kappa v\,.##
 
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1. What is Schur's Lemma and why is it important?

Schur's Lemma is a fundamental result in algebra that states that any finite-dimensional irreducible representation of a group is either one-dimensional or equivalent to a direct sum of copies of the trivial representation. This lemma is important because it provides a powerful tool for understanding the structure of group representations and has applications in various areas of mathematics such as group theory, representation theory, and linear algebra.

2. How is Schur's Lemma proven?

Schur's Lemma can be proven using a combination of techniques from linear algebra, representation theory, and group theory. The proof typically involves showing that any non-trivial endomorphism of an irreducible representation must be a scalar multiple of the identity map, and thus the representation is either one-dimensional or equivalent to the trivial representation.

3. Is there a simple proof of Schur's Lemma?

There is no universally agreed upon definition of what constitutes a "simple" proof, but there are several different approaches to proving Schur's Lemma that may be considered more straightforward or elegant than others. These include using basic properties of group representations, character theory, and the representation theory of finite groups.

4. What are the applications of Schur's Lemma?

Schur's Lemma has a wide range of applications in mathematics, including in the study of finite groups, Lie algebras, and algebraic number theory. It is also used in physics, particularly in the study of quantum mechanics and group theory.

5. Are there any generalizations or extensions of Schur's Lemma?

Yes, there are several generalizations and extensions of Schur's Lemma, including the Brauer-Wall Group Theorem, which applies to infinite-dimensional representations, and the Schur-Weyl duality, which relates representations of a group to representations of a symmetric group. There are also variants of Schur's Lemma for different types of groups, such as compact groups and Lie groups.

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