Is a Lie Algebra inherently associated with a topology?

Question

For context, at work I'm part of a math hobby group that recently studied Lie groups and Lie algebras. There's a big result the correspondence between Lie groups and Lie algebras that we covered a few weeks ago. I don't understand precisely why the exponential map lets you map back and forth between Lie groups (in a neighborhood around a point I think) and Lie algebras. I have a vague visual intuition about wrapping a sheet of paper around a sphere-without-the-antipode, but nothing more than that. My goal is to understand that result, but as a first step I want to understand Lie algebras a little bit better independently of their association with Lie groups. I also don't understand Lie groups very well, and I'm hoping that understanding Lie algebras first and then understanding the correspondence is a productive way to build an understanding of Lie groups.

With that being said, here's the definition of a Lie algebra paraphrased from Wikipedia.

It's a vector space $\mathfrak{g}$ with field of scalars $F$ equipped with a binary operation $[\cdot, \cdot]$, the Lie bracket, that is bilinear, anticommutative, and satisfies the Jacobi identity.

I think of the motivating example of a Lie algebra as $(\mathbb{R}^3, \times)$.

$(\mathbb{R}^3, \times)$ has an obvious topology associated with it, the standard topology on $\mathbb{R}^3$.

However, the definition of a Lie algebra on Wikipedia does not insist that the underlying vector space is a topological vector space.

For concreteness, I'm trying to understand, just for right now, why we don't insist that the underlying vector space in a Lie algebra is endowed with a topology.

Answer 1

Because Lie algebras are defined over any field, not just $\Bbb R$ or $\Bbb C$. If, say, $F$ is a field with three elements and you define, say, the Lie algebra$$\mathfrak{sl}(3,F)=\left\{\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\,\middle|\,a,b,c,d,e,f,g,h,i\in F\wedge a+e+i=0\right\},$$then the underlying vector space is a $8$-dimensional $F$-vector space. What's the point of defining a topology in it that turns it into a topological vector space?

Besides, even if we are working over $\Bbb R$ or $\Bbb C$, then, assuming that $\mathfrak g$ is finite-dimensional, it always has a natural topology, which is the only topology induced on it by any norm defined on $\mathfrak g$.

Answer 2

A $n$-dimensional vector space over $\mathbb{R}$ (or $\mathbb{C}$) always has an unique (Hausdorff) topology isomorphic to the standard topology on $\mathbb{R}^n$ (or $\mathbb{C}^n$).

So insisting that the underlying vector space in a Lie algebra is endowed with a topology gives you nothing new.