Left invariant vector fields and Lie brackets of the upper triangular matrices

analysislie-algebrassmooth-manifoldsVector Fields

This question was asked in my assignment on smooth manifolds and I am not able to solve it because vector fields has been 1 of my weak points. I have been following introduction to smooth manifolds by Lee along with my notes.

Question: Consider the group H of upper triangular matrices of the form $\begin{bmatrix} 1&a&c \\ 0& 1& b\\ 0& 0& 1\end{bmatrix}$ with the matrix multiplication being the group operation. Show that H is a Lie Group. Characterise the left invariant vector fields of H and describe the Lie bracket on the Lie algebra of H.

Attempt: I have proved that H is a lie group.

I am having troubles in using the definition of left invariant vector fields:

On pg 189 of John Lee the definition is given as: Suppose G is a Lie group. Recall that G acts smoothly and transitively on itself by left translation: $L_g(h) = gh$. A vector field X on G is said to be left-invariant if it is invariant under all left translations, in the sense that it is $L_g$ -related to itself for every $g \in G.$ More explicitly, this means that $d(L_g)_{g'}(X_{g'}) =X_{gg'}$ for all $g,g' \in G$.

Trying to use the definition the problem I am facing is that I am not sure how should I define $X_{g'}$ and ${X_{gg'}$ Can you please help me with that

Describing Lie brackets on the Lie Algebra of H:

A Lie algebra (over R) is a real vector space $g$ endowed with a map called the bracket from $g \times g \to g$, usually denoted by $(X,Y) \to [X,Y]$ , that satisfies the following properties for all X,Y,Z $\in g$:(i) Bilinearity , Anti-symmetry, Jacobi Identity and one properties related to smooth functions.

Lie bracket is an operator $[X,Y] : C^{\infty}(M) \to C^{\infty}(M)$ is called a Lie bracket of X and Y. $[X,Y]f = XYf -YXf$.

The set of all upper triangular matrices is a vector space wrt +. Now, I think the operator will be if X,Y are Upper triangular matrices and f another matrix which is upper triangular then the operator will be $XYf -YXf$ with usual matrix multiplication. Have I defined the lie bracket correctly? It will satisfy all the properties of Lie brackets correctly.

Can you please verify the claim related to Lie Brackets and tell me how should I approach the part related to vector fields as I am not very comfortable in that.

Best Answer

When a vector field on $H$ is left-invariant, it is completely determined by its value at the identity of $H$. This is an immediate consequence of the definition.

The tangent space at the identity of $H$ is isomorphic to $\mathbb R^3$ as a vector space, because $H$ is three-dimensional. A basis is given by the matrices $$ E_1 = \left( \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right), \quad E_2 = \left( \begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right), \quad E_3 = \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right).$$ Now computing $(L_g)_*E_i$ and taking linear combinations gives all possible left-invariant vector fields on $H$.

The commutator of two vector fields $X$ and $Y$ is defined as $[X,Y] = XY-YX$. Here you can easily compute \begin{align*} [E_1,E_2] & = E_1E_2-E_2E_1 = 0, \\ [E_1,E_3] & = E_1E_3-E_3E_1 = E_2 \\ [E_2,E_3] & = E_2E_3-E_3E_2 = 0. \end{align*} Since $(L_g)_*([X,Y]) = [(L_g)_*X,(L_g)_*Y]$, again the values taken by the commutators on $H$ are all determined by the above three identities. This operation gives the structure of Lie algebra to the space of left-invariant vector fields.

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