# About Schur's lemma in lie algebra

• A
• HDB1
In summary, Schur's Lemma states that a matrix whose determinant is zero is invertible. This can be used to determine the inverse of a matrix from its determinant.
HDB1

Let $\phi: L \rightarrow g I((V)$ be irreducible. Then the only endomorphisms of $V$ commuting with all $\phi(x)(x \in L)$ are the scalars.

Could you explain it, and please, how we can apply this lemma on lie algebra ##L=\mathfrak{s l}(2)##thanks in advance,

Dear @fresh_42 , If you could help, I would appreciate it, thanks in advance,

HDB1 said:

Let $\phi: L \rightarrow g I((V)$ be irreducible. Then the only endomorphisms of $V$ commuting with all $\phi(x)(x \in L)$ are the scalars.

Could you explain it, and please, how we can apply this lemma on lie algebra ##L=\mathfrak{s l}(2)##thanks in advance,
It says if ##\varphi \, : \,V\longrightarrow V## is a linear transformation of the vector space ##V## and ##\phi\, : \, L \rightarrow \mathfrak{gl}(V)## an irreducible representation of the Lie algebra ##L## then
\begin{align*}
[\phi(X),\varphi ](v)&=(\phi(X)\cdot \varphi -\varphi \cdot \phi(X))(v)=\phi(X).\varphi (v)-\varphi (\phi(X).v)=0 \text{ for all }X\in L\\ &\Longrightarrow \\
\varphi(v)&=\lambda \cdot v\text{ for some }\lambda \in \mathbb{K}
\end{align*}

Note:
a) ##\varphi \in \operatorname{End}(V)=\mathfrak{gl}(V)##
b) ##\{\alpha \in \mathfrak{gl}(V)\,|\,[\alpha,\beta]=0\text{ for all }\beta\in \mathfrak{gl}(V)\}=Z(\mathfrak{gl}(V)).##
c) Schur's lemma can therefore be phrased as follows:

A linear transformation ##\varphi ## of a finite-dimensional representation space ##V## of an irreducible representation ##\phi## of a Lie algebra ##L,## i.e. ##V## is an irreducible ##L## module, that lies in the center of ##\mathfrak{gl}(V)## is a scalar multiple of the identity matrix.

Consider the case ##L=\mathfrak{sl}(2)\, , \,V_2=\mathbb{K}^2## with ##x.v=[x,v]## being the Lie multiplication of ##\mathfrak{sl}(2)\ltimes V_2## we have seen before, will say: ##x.v## is the multiplication of a matrix ##x\in \mathfrak{sl}(2)## and a vector ##v\in V_2.## This is an irreducible representation, since ##V_2## has no one-dimensional submodule ##U=\mathbb{K}u## such that ##x.u \in U## for every ##x\in \mathfrak{sl}(2).## (Prove it!)

So all conditions of Schur's lemma are fulfilled. Now, if we have a matrix ##\varphi = A= \begin{pmatrix}a&b\\c&d\end{pmatrix}\in \mathfrak{gl}(V_2)## such that
$$0=[\phi(X),A]=[X,A]=X\cdot A- A\cdot X\text{ for all } X\in \mathfrak{sl}(2) \;\;\Longleftrightarrow \;\;AX=XA$$
then ##A=\lambda \cdot \begin{pmatrix}1&0\\0&1\end{pmatrix}## for some ##\lambda \in \mathbb{K}.##

You can check this yourself. Prove:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\cdot \begin{pmatrix}h&x\\y&-h\end{pmatrix}=\begin{pmatrix}h&x\\y&-h\end{pmatrix}\cdot \begin{pmatrix}a&b\\c&d\end{pmatrix} \text{ for all }x,h,y\;\;\Longrightarrow \;\;b=c=0 \text{ and }a=d$$

HDB1

## What is Schur's lemma in Lie algebra?

Schur's lemma is a fundamental result in Lie algebra, which states that any endomorphism of a simple Lie algebra must be either zero or an isomorphism. It is named after the mathematician Issai Schur, who first proved the lemma in 1905.

## What is the significance of Schur's lemma in Lie algebra?

Schur's lemma is a powerful tool in the study of Lie algebras. It helps to identify and classify simple Lie algebras, which are the building blocks of more complex Lie algebras. It also has applications in representation theory and algebraic geometry.

## What is the difference between Schur's lemma in Lie algebra and Schur's lemma in group theory?

While both lemmas are named after Issai Schur and have similar statements, they apply to different mathematical structures. Schur's lemma in Lie algebra deals with endomorphisms of Lie algebras, while Schur's lemma in group theory deals with endomorphisms of groups.

## How is Schur's lemma used in the proof of the Cartan-Killing criterion?

The Cartan-Killing criterion is a theorem that characterizes the structure of semisimple Lie algebras. Schur's lemma is used in the proof of this criterion to show that the Killing form, a symmetric bilinear form on a Lie algebra, is non-degenerate.

## Is Schur's lemma applicable to all Lie algebras?

No, Schur's lemma only applies to simple Lie algebras. For non-simple Lie algebras, there are more general versions of Schur's lemma, such as the Jacobson-Morozov lemma.

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