Lie Groups and Lie Algebras Lecture 6 notes

Left invariant vector fields and one-parameter subgroups: Examples and Consequences

Summary (review):
The space of left (right) invariant vector fields is isomorphic to the tangent space $T_e G$ at the identity (because each invariant vector field corresponds to a tangent vector of $e$ )
- Left (right) invariant vector fields are complete (flow can be defined on $\mathbb{R}$ )
- Integral curves of left (right) invariant vector fields through the identity are one-parameter subgroups
- The flow on $G$ generated by a left invariant vector field $\xi$ can be written in the form $\phi_{\xi}^t(a) = ae^{t \xi_0}$
- Any left invariant vector field commutes with any right invariant vector field
- The space of left invariant vector fields is closed under the Lie Bracket

Important remark:
Let $G \subseteq GL(n, \mathbb{R})$ be a matrix Lie group. Then, a tangent vector to $G$ at $X_0 \in G$ can (and will) be considered as a certain $n \times n$ matrix.
We use the idea that tangent vectors to $G$ are tangent vectors to curves lying in $G$ .
Then, a curve passing through $X_0$ is a family of matrices $X(t) \in G$ (smooth), with $X(0) = X_0$ , so its derivative wrt $t$ is an $n\times n$ matrix:
$X(t) = x_{ij}(t)$ for $i, j \in (1, n)\,\implies dX(0)/dt = x'_{ij}(0)$ for $i, j \in (1, n)$ , which is an element of $T_{X_0}G$

Theorem: Tangent Space of Gl(n, R)

Proposition: The tangent space $T_X G$ is a certain subspace in $M_{n, n}$ (the space of all matrices), which depends on $X$ . However, if $G = GL(n, \mathbb{R})$ , then the tangent space $T_X GL(n, \mathbb{R})$ coincides with $M_{n, n}$ at each point $X$ .
Proof:
By definition, every $A \in GL(n, \mathbb{R})$ has a nonzero determinant (i.e. it is invertible).
Choosing some norm on matrices $M_n(\mathbb{R})$ , the function $\text{det}: M_n(\mathbb{R}) \to \mathbb{R}$ is continuous, so there is $\epsilon > 0$ s.t. for all matrices $H$ with $||H|| < \epsilon$ , $\text{det}(I + H) \neq 0$ . #-> $I$ has determinant 1, so there exists some neighborhood with nonzero determinant
Thus, for all matrices $X$ , there exists $\eta > 0$ s.t. for $-\eta < t < \eta$ , $\text{det}(I + tX) \neq 0$ . #-> For some choice of $X$ , we can choose $\eta$ s.t. for any $-\eta < t < \eta$ , $tX$ has nonzero determinant
This means that the path $P(t) = I + tX$ is a path with derivative $X$ and with $P(t) \in GL(n, \mathbb{R})$ for sufficiently small $t$ , so any matrix $X$ can be a tangent matrix of some path on $GL(n, \mathbb{R})$ .

Theorem: Left Invariant Vector Field On Gl(n, R)

Proposition: Take an arbitrary $A \in M_{n,n}$ as a tangent vector to $GL(n, \mathbb{R})$ at $E = \text{id}$ . Then, the corresponding left invariant vector field on $GL(n, \mathbb{R})$ is $\xi(X) = XA$ .
Proof: By construction, $\xi(X) = dL_x(A) = XA$ (left multiplication by $X$ is a linear map, so it coincides with its own differential)

Theorem: One Parameter Subgroup of Gl(n, R)

Proposition: Take $A \in M_{n,n}$ as a tangent vector of $GL(n, \mathbb{R})$ . Then, the corresponding one-parameter subgroup is given as $\text{exp}(t, A) = e^{tA} = E + tA + t^2\frac{A^2}{2!} + \dots$ #-> Taylor series expansion with matrix exponential
Proof:
It is left as an exercise to check that this series converges for any $A$ and $t$ absolutely (and uniformly on any interval $t \in (-T, T)$ ).
1) Moreover, the resulting matrix function is smooth wrt $t$ , and $\frac{d}{dt} e^{tA} = Ae^{tA} = e^{tA}A$ .
2) Besides, $e^{(t + s)A} = e^{tA} e^{sA}$ . (Note that $e^A e^B != e^{A + B}$ if $AB \neq BA$ )
Then, 1) means that $e^{tA}$ is an integral curve of $\xi(X) = XA$ through the identity and 2) means that $e^{tA}$ is a one-parameter subgroup (with initial tangent vector $A$ ).

The above two statements hold for any Lie subgroup $G \subseteq GL(n, \mathbb{R})$ . #-> Which statements? Future Dylan Note: I'm dumb, it's the above two numbered statements
This follows from the uniqueness condition: for any tangent vector $\xi_0 \in T_e G$ , there is a unique one-parameter subgroup $f: \mathbb{R} \to G$ s.t. $\frac{d}{dt}f(0) = \xi_0$ and there is a unique left invariant vector field $\xi$ s.t. $\xi(e) = \xi_0$ .
Corollary: Let $g = T_E G \subseteq M_{n,n}$ be the tangent space of a matrix group $G \subseteq GL(n, \mathbb{R})$ at the identity. Then, the tangent space at any other point $X \in G$ is given as $T_X G = Xg = gX$
Corollary: Considering a system of linear ODE $\frac{d}{dt}\mathbf{x} = A(t)\mathbf{x}$ , $\mathbf{x} \in R^n$ , letting $A(t) \in G$ for any $t \in \mathbf{R}$ , then the fundamental solution $X(t)$ belongs to $G$ for any $t \in \mathbf{R}$ (recall that $X'(t) = A(t)X(t)$ and $X(0) = E$ ).

For matrix groups, $\xi(x) = dL_X(A) = XA$ , where $X \in G$ and $A \in T_E G$

Definition: Exponential Map

The exponential map $\text{exp}: T_e G \to G$ is defined by $\text{exp}(\xi_0) = \text{exp}(t\xi_0) |_{t=1}$ , where $\xi_0 \in T_e G$ and $\text{exp}(t\xi_0)$ denotes the one-parameter subgroup in $G$ with the initial vector $\xi_0$ .
.#-> This looks like a tautology, but in reality we are defining a map from the tangent space to the group by taking the one-parameter subgroup generated by a tangent vector and evaluating it at t=1.
Remark:
Our notation $\text{exp}(t\xi_0)$ can now be understood in two different ways: as the image under $\text{exp}$ of the tangent vector $t\xi_0$ , or as the point on the one-parameter subgroup $\text{exp}(t\xi_0)$ with parameter $t$ .
In fact, these points coincide, so the notation causes no confusion.

Theorem: Properties of the Exponential Map

Properties of the exponential map:
1) $\text{exp}$ is smooth and globally defined on $T_e G$ as a whole
2) the differential of $\text{exp}$ at zero is the identity operator: $d \text{exp}: T_e G \to T_e G$ , $d \text{exp}(\xi_0) = \xi_0$
3) $\text{exp}$ is a local diffeomorphism at a neighborhood of zero
Proof for 2):
notice that $\text{exp}(0) = e$ , where $0 \in T_e G$ is the zero vector. Then, $d \text{exp}|_{0}: T_0(T_e G) \to T_e G$ .
Notice that $T_e G$ is a vector space and therefore the tangent space to it can be identified with $T_e G$ itself. That is, $T_0(T_e G) = T_e G$ .
Recall that for $f: M \to N$ , its differential $df|_{P}: T_P M \to T_{f(P)} N$ is defined by $df|_{P}(\gamma'(0)) = \frac{d}{dt}|_{t=0} f(\gamma(t))$ , where $\gamma(0) = P$ .
In our case, for $\xi \in T_e G$ , we set $\gamma(t) = t\xi \in T_e G$ , which means $\gamma'(t) = \xi$ . Then, $d \text{exp}|_{0}(\xi) = d \text{exp}|_{0}(\gamma'(0)) = \frac{d}{dt} |_{t=0} \text{exp}(t\xi) = \xi$ .
That is, $d \text{exp}|_{0} = \text{id}$ .

Left translation allows us to identify the tangent space $T_x G$ at an arbitrary point $x \in G$ . This identification is natural, and implies the following topological property of Lie Groups:

Theorem: Lie Group Parallizability

Any Lie group $G$ is parallelizable, i.e. its tangent bundle $TG$ is trivial: $TG \cong G \times \mathbb{R}^n$ (with $n = \text{dim } G$ ). #-> The above implies that every point's tangent space is isomorphic to one another, so the tangent bundle is just the group times the tangent space at the identity
Proof:
In general, the triviality of the bundle $TM$ means that there is a smooth map $\phi: M \times \mathbb{R}^n \to TM$ which is linear on each fiber
i.e. $\phi(x, \xi) = (x, A_x(\xi))$ , where $A_x: \mathbb{R}^n \to T_x M$ is a linear isomorphism #-> This is straightforward: if this condition is true, then the manifold should be isomorphic to the product
In our case, such a map is given by $\phi: G \times \mathbb{R}^n \to TG$ by $\phi(x, \xi_0) = (x, dL_x(\xi_0)), \xi_0 \in T_e G = \mathbb{R}^n$ #-> $dL_x(\xi_0)$ sends a tangent of $T_e G$ to a tangent of $T_x G$
Corollary:
- Any Lie group $G$ is orientable #-> orientability: ability to assign an orientation to each side (e.g. not a mobius strip)
- Note that not all orientable manifolds are Lie groups (e.g. the $S^2$ sphere)
- Among closed 2-dimensional surfaces, only the torus $T^2$ may carry the structure of a Lie group #-> That's insane! bc only the torus is parallelizable
- Alternative definition for parallelizable: existence of smooth vector fields $\{V_1, \dots, V_n\}$ s.t. for any $p \in M$ the tangent vectors $\{V_1(p), \dots, V_n(p)\}$ form a basis of $T_p M$
- Then, the 2-sphere is obviously not parallelizable, as the hairy ball theorem states that there exist no smooth vector fields on $S^2$ . Thus, the sphere is not a Lie group. #-> interesting proof, seems a bit backwards lol

Relationship between Lie groups and Lie algebras (explicit):
Let $G \in GL(n, \mathbb{R})$ be a matrix Lie group, and let $A, B \in T_E G$ and $\xi(X) = XA, \eta(X) = XB$ be corresponding left invariant vector fields on $G$ .
Then, the Lie bracket $[\xi, \eta]$ is the left invariant vector field of the form $X(AB - BA)$ .
.#-> by def of Lie bracket, $[\xi, \eta](X) = \xi(\eta(X)) - \eta(\xi(X))$ . In our case, = $(XBA - XAB) = X(BA - AB)$ #-> shouldn't this be $X(AB - BA)$ ? wait nvm

Theorem: Corollary: Tangent Space Closed Under Matrix Commutator

The tangent space $T_E G$ at the identity of any matrix Lie group is closed under the matrix commutator $[A, B]: A, B \mapsto AB - BA$
Proof:
Notice that it suffices to prove $[\xi, \eta] = X(AB - BA)$ for $GL(n, \mathbb{R})$ only, then it will hold for any Lie subgroup $G \subseteq GL(n, \mathbb{R})$ automatically.
In local coordinates, the proof is "straightforward": $[\xi, \eta]_{ij} = \sum_{k, l}(\xi_{kl} \frac{d}{d x_{kl}} \eta_{ij} - \eta_{kl} \frac{d}{d x_{kl}} \xi_{ij})$ #-> not very straightforward...
We have $\xi_{kl} = \sum_{\alpha} x_{k \alpha}a_{\alpha l}$ and, similarly, $\eta_{kl} = \sum_{\alpha} x_{k \alpha}b_{\alpha l}$ .
Here by $\xi_{ij}, x_{ij}, a_{ij}$ , etc. we denote the matrix coefficients of $\xi, X, A$ , etc.
Hence, $[\xi, \eta]_{ij} = \sum_{k,l}(\sum_{\alpha} x_{k \alpha}a_{\alpha l} \frac{d}{d x_{kl}} (\sum_{\beta} x_{i \beta} b_{\beta j}) - \dots)$
Only when $(kl) = (i \beta)$ does the partial derivative not vanish, and in this case it equals $b_{\beta j}$ . Hence, we can replace $k$ by $i$ and $l$ by $\beta$ , and we get
$= \sum_{\alpha,\beta}(x_{i \alpha}a_{\alpha \beta}b_{\beta j} - x_{i \alpha}b_{\alpha \beta}a_{\beta j})$
$= \sum_{\alpha, \beta}x_{i \alpha}(a_{\alpha \beta}b_{\beta j} - b_{\alpha \beta}a_{\beta j})$
this formula means exactly that $[\xi, \eta](X) = X(AB - BA)$ as stated above.