I am studying Lie groups and I am aware that we can think of a Lie group as a differentiable manifold which means we can talk about functions, curves and tangent spaces.

For an abstract manifold $M$, if we have an open neigbourhood about a point $p \in M$ we can define a chart $\phi: M \rightarrow \mathbb{R}^n $. We define the tangent space at the point $p$ as the vector space of directional derivative operators to curves passing through the point $p$ which act on functions $f:M \rightarrow \mathbb{R}$. In the chart $\phi$ with coordinate functions $\{x^i\}$, I can define tangent vector to the curve $\gamma$ at $p$, given by $V$, as

$$ V(f) = \frac{d}{dt}(f \circ \gamma) \bigg|_{t=0} \\ = \frac{dx^i}{dt}\bigg|_{t=0} \frac{\partial (f \circ \phi^{-1})}{\partial x^i}\bigg|_{\phi(p)} \\ = \frac{dx^i}{dt}\bigg|_{t=0} \frac{\partial}{\partial x^i}\bigg|_p f. $$

So we make the identification that the vector is given by $$ V = \frac{dx^i}{dt}\bigg|_{t=0} \frac{\partial}{\partial x^i}\bigg|_p. $$

Now in my studies of Lie algebras I have been introduced to the idea that the Lie algebra of a Lie group $G$ is the tangent space at the identity $T_e(G)$ equipped with a bracket. In my lectures we were told that in order to make this Lie group to Lie algebra transition, it's easiest to use matrices and we perform the mapping from $G \rightarrow GL(n,F)$ as

$$ \frac{\partial}{\partial x^i} \mapsto \frac{\partial g(t)}{\partial x^i}, $$

where $g(t)$ is some curve through the matrix Lie group.

I have some questions regarding this:

  1. In order to take the derivative of elements $g(t)$ of the Lie group, do we have to assume that it's also a vector space because we need the notion of addition and scalar multiplication?
  2. How do I know that this is in the tangent space of the matrix group?
  3. How do these basis vectors act on functions? i.e. above I can say that the abstract basis vectors act on functions as $$ \frac{\partial}{\partial x^i}\bigg|_{p} f = \frac{\partial (f \circ \phi^{-1})}{\partial x^i}\bigg|_{\phi(p)}.$$ What would $ \frac{\partial g(t)}{\partial x^i}\bigg|_p f $ be?

1 Answer 1

  1. Indeed, every group whose elements are $n\times n$-matrices is a subset of the linear space $\mathbb{R}^{n^2}$. One can thus consider sums and linear combinations of the group elements. It is important to note that a general tangent vector is not a group element.

  2. It is, by definition, in the tangent space. As commented above, this does not meant that a tangent vector is in the Lie group.

  3. You can always think of the tangent space to a matrix group as a linear subspace of $\mathbb{R}^{n^2}$. Similarly, the matrix group is a submanifold of $\mathbb{R}^{n^2}$. finally, tangent vectors act on functions exactly as they always do in Euclidean space or submanifolds thereof.

  • $\begingroup$ Thank you for your answers. Regarding question two, I haven't studied much DG as I have a physics background. I've only learned differential geometry from an intrinsic point of view, so the idea of a tangent vector for me requires some function $f$ to act upon in order to define it like i defined in my question. The idea that I can take the derivative of a curve $g(t)$ without a function is unfamiliar to me. Does this arise because when we make an isomorphism between an abstract group and a matrix group we are now able to perform operations on curves which we weren't able to before? $\endgroup$ May 17, 2018 at 19:27
  • 1
    $\begingroup$ There are a few different ways to think of tangent vectors. My answer works best with the definition using equivalence classes of curves. You can find this definition in many textbooks. $\endgroup$ May 17, 2018 at 20:24

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