Clarification on Lie Algebras Notes: "Let $\mathfrak{gl}(V) = End(V)$"

Question

I wanted to ask about something confusing in my Lie Algebras notes.

At one point we define the Lie Algebra $\mathfrak{gl}(V)$ to be the vector space $GL_n(V)$ with the bracket $[x , y ] = xy - yx$. (Here $V$ is an $n$-dimensional vector space over a field, $k$.)

Later, it says "let $\mathfrak{gl}(V) = End(V)$ with bracket given by $[x,y] = xy - yx$.

The use of the word "let" here confuses me. Given the same notation of $\mathfrak{gl}(V)$ I expect that these two Lie Algebras should be equivalent.

However, my understanding is that $GL_n(V)$ is the vector space of invertible $n \times n$ matrices over the field $k$, with respect to a chosen basis of $V$, and that $End(V)$ is simple the vector space of Endomorphisms of $V$.

However, this definition of $\mathfrak{gl}(V)$ seems to imply that, as vector spaces, $End(V) = GL_n(V)$, but this then implies that everything in $End(V)$ is invertible, which is not the case.

I am very confused by this. Should the $End(V)$ be replaced with $Aut(V)$? Am I misunderstanding something simple about Lie Algebras, $GL_n(V)$ or $End(V)$?

I would really appreciate some help understanding this, thank you.

Answer 1

The first reference to $GL$ is wrong and the reference to $End$ is correct. Indeed, $GL(V)$ is not preserved by the bracket since for instance a matrix bracketed with itself is not invertible.

Answer 2

There is a misteka from the start: $\mathfrak{gl}(V)$ is the vector space of all endomorphisms of $V$. In fact, $GL_n(V)$ isn't even a vector space.

And the bracket operation on $\operatorname{End}(V)$ should read: $[x,y]=x\circ y-y\circ x$.

Answer 3

Complementary info, where the confusion may come from: uppercase letters are used to denote Lie groups, while lowercase letters are used to denote their associated Lie algebras. $GL_n(V)$ (not a vector space) is the Lie group of invertible $n\times n$ matrices from End$(V)$, while $gl_n(V)$ is its associated Lie algebra (the vector space End$(V)$ along with the commutator). If instead you pick the Lie group $SL_n(V)$ of invertible matrices with determinant 1, then the associated Lie algebra $sl_n(V)$ is composed of those matrices with trace $0$ (and the commutator as product).