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

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.

• Is $GL_n(V)$ a vector space? – Hugo Oct 10 '18 at 9:50

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.

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: $$[x,y]=x\circ y-y\circ x$$.

• Thank you for the answer and clarification on the bracket. The bracket also confused me because I wasn't sure what composition or multiplication meant in the vector space $\mathfrak{gl}(V)$. Is it correct to say that $\mathfrak{gl}(V)$ satisfies all the conditions of being a vector space, but ALSO has another operation other than addition defined for it? – user366818 Oct 10 '18 at 9:59
• @user366818 Indeed. Besides addition, $\mathfrak{gl}(V)$ also has the composition operation. – José Carlos Santos Oct 10 '18 at 10:00

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).

• Thank you, this makes it clear that although $\mathfrak{gl}(V)$ is the associated Lie Algebra of $GL_n(V)$, they do not have the same elements, which is an assumption I had made. – user366818 Oct 10 '18 at 10:02
• @user366818 The usual interpretation is that the Lie algebra is the tangent space of the (connected component of the identity element of the) Lie group. So, although a group is not a vector space, its Lie algebra indeed is, with vectors pointing to local directions inside the group (but enriched with a new interesting product for vectors, the commutator). I tend to visualize a sphere with a plane on top... – Jose Brox Oct 10 '18 at 11:50
• Thank you for the explanation. In the course I am taking I believe we are trying to define everything strictly algebraically, and so I believe we are avoiding this interpretation that the Lie algebra is the tangent space of the Lie group. – user366818 Oct 10 '18 at 11:52