NOTES ON STATISTICS, PROBABILITY and MATHEMATICS

Lie Group and Lie Algebra for \(SL(2, \mathbb C)\)

From here.

The group is a set

\[SL(2,\mathbb C) =\left\{\begin{bmatrix}a & b \\ c & d\end{bmatrix} \in \mathbb C^4 \mid ad - bc = 1 \right\}\]

with the group operation of matrix multiplication.

This set has a topological space related to \(\mathbb C.\)

To check that it is also a topological manifold we need to build charts:

\[\mathcal U =\left\{\begin{bmatrix}a & b \\ c & d\end{bmatrix} \in SL(2, \mathbb C) \mid a \neq 0 \right\}\]

so that the chart map \(x\)

\[x: \mathcal U \to \mathbb C^3\]

because there are only \(3\) degrees of freedom, since the determinant has to be \(1.\)

This is defined as

\[x\left( \begin{bmatrix}a & b \\ c & d\end{bmatrix}\right)=(a,b,c)\]

with the inverse being

\[x^{-1}(a, b, c) = \begin{bmatrix}a & b \\ c & \frac{1 + bc}{a}\end{bmatrix}\]

both are continuous, forming a homeomorphism.

Another chart can be

\[\mathcal V =\left\{\begin{bmatrix}a & b \\ c & d\end{bmatrix} \in SL(2, \mathbb C) \mid d \neq 0 \right\}\]

with

\[y\left( \begin{bmatrix}a & b \\ c & d\end{bmatrix}\right)=(b,c,d)\]

and

\[y^{-1}(b, c, d) = \begin{bmatrix}\frac{1 + bc}{d} & b \\ c & d\end{bmatrix}\]

The next step is to check the chart compatibility.

The charts constitute an atlas \(\mathcal A,\) which together forms a differentiable manifold.

So \(SL(2,\mathbb C)\), equipped with a topology and atlas \(\left( \mathcal O, \mathcal A \right)\) is a Lie group since both the map afforded by the matrix operation and the inverse are smooth.

The Lie Algebra \(\frak{sl}(2, \mathbb C)\)

A Lie algebra is the set of left-invariant vector fields (section of the tangent bundle) on the group \(G\)

\[\frak{L}(G)=\left\{X \in \Gamma(TG) \mid \left( l_{g *} X_h \right)_{[l_g h = gh]} = X_{gh} \right\}\]

where the left translation \(l_g\) is defined as

\[l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}}: SL(2, \mathbb C) \to Sl(2, \mathbb C)\]

defined as

\[l_{\tiny \begin{bmatrix}a & b \\ c & d\end{bmatrix}}\begin{bmatrix}e & f \\ g & h\end{bmatrix}=\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}e & f \\ g & h\end{bmatrix}\]

and \(l_{g *}\) is the push-forward of the map \(l_{g}\), in this case on the vector field \(X\) at some point \(h \in G\), the result of which will be at \(l_g h = gh\).

An important result, derived in another video by Prof Schuller (in here), is that there is an isomorphism between the tangent space at the identity to the group

\[T_{\tiny \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}} \, SL(2, \mathbb C) \equiv L(2, \mathbb C)\]

with \(L\) being the set of left-invariant vector fields (the defining set in the Lie algebra).

So rather than looking at vector fields over the manifold, it is enough to look at the corresponding one vector at the identity.

In vector fields the bracket (commutator) is defined: one vector field, acting on another vector field, acting on a function. Now we need a Lie bracket at the tangent plane at the identity as:

\[[\cdot, \cdot]: T_{id}SL(2, \mathbb C) \times T_{id}SL(2, \mathbb C) \to T_{id}SL(2, \mathbb C)\]

\[[A, B] = l_{g^{-1}*}[l_{g*}A,l_{g*}A]\]

using the push-forward of the left translation where \(A \in T_{\tiny \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}SL(2, \mathbb C).\)

The calculation proceeds as follows:

\(g \in SL(2, \mathbb C)\) so the push-forward of the left-translation by this \(g\) will act on an element at the identity:

\[\left [l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}*} \left(\frac{ \partial }{\partial x^i} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right ]_{\tiny \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\; f = \left [l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}*} \left(\frac{ \partial }{\partial x^i} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right ]_{\tiny \begin{bmatrix}a & b \\ c & d\end{bmatrix}}\; f \]

To do so we need a chart, and the \(\mathcal U\) is the chart that contains the identity. The coordinates afforded by the chart at the identity are \(x^i\) with \(i = 1,2,3.\) So pushing forward the tangent vector at the identity to the point \(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\) to get a tangent vector at this point (which happens to just be \(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\) in this case). This is acting on a function \(f.\) So we take the \(i\)-th tangent vector at the identity in the \(\mathcal U\) chart (coordinate system) getting a tangent vector at the point \(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\). And since this point is any point in the chart, we can generate the entire left-invariant vector field from one single vector at the identity.

The push-forward generates a vector field by the \(i\)-th tangent vector at the identity \(\left(\frac{ \partial }{\partial x^i} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\) after the map underlying the push-forward, \(l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}}\), acting on \(f\).

\[\left [l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}*} \left(\frac{ \partial }{\partial x^i} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right ]_{\tiny \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\; f = \left(\frac{ \partial }{\partial x^i} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}} \left(f \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \right) = \left.{\partial_i \left( f \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right)}\right|_{x\left(\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)}\]

Notice that in the intermediate expression Prof Schuller is invoking the definition of the push-forward of a vector field \(\frac{\partial}{\partial x^i}\). His explanation of the pushforward can be found here, where he defines as \(\phi^*(X)f := X(f \circ \phi),\) where \(\phi\) is a smooth map between manifolds \(M\) and \(N.\) The function \(f\) is in the target \(N,\) while \(X \in T_pM,\) and \(\phi^*(X) \in T_{\phi(p)}N.\) The RHS of the definition says, that you take the vector \(X\) and act on \(f\) after \(\phi.\) Given that \(\phi\) takes you to \(N\) from \(M\), i.e. $M N $ and \(f\) takes you to the real numbers, we are providing \(X\) with a differentiable function \(M \to N \to \mathbb R.\) This makes sense.

Yes, he considers this expression, \(\frac{\partial}{\partial x^i}\), to be the vector field, but it doesn’t sound correct. Then he says that the pushforward of a vector field \(\frac{\partial}{\partial x^i}\) at \(\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\) acting on \(f\) after the smooth map underlying the pushforward, \(lg_{\tiny \begin{bmatrix}a & b \\ c & d \end{bmatrix}}\). However, \(\frac{\partial}{\partial x^i}\) at \(\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\) doesn’t sound like a vector field. He does mention that \(\frac{\partial}{\partial x^i}\) is not a partial derivative: it uses the coordinates \(x^i\) and only becomes a partial derivative after applying \(x^{-1}\). This is also explained here.

The last expression meant to make the partial derivative work. In the expression $ i ( f l{ \[\begin{bmatrix}a & b \\ c & d\end{bmatrix}\]

} x^{-1})$ the \(x\) is in the chart \(\mathcal U,\), i.e. \((\mathbb C^*, \mathbb C, \mathbb C)\), where the asterisk here means that it’s the complex plane minus \(0\). This inverse is going to a point in the manifold, where it gets left translated, i.e. \(l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}},\) at which point the function \(f\) takes you to the complex numbers \(\mathbb C.\) So the derivative with respect to the first, second and third entry of \((\mathbb C^*, \mathbb C, \mathbb C)\).

With the usual trick

\[=\partial_i \left( f \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right) = \partial_i \left.{ \left(\left( f \circ x^{-1}\right) \circ \left(x \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right)\right)}\right|_{x\left(\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)}\]

the last bit \(x \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\) goes from \(\mathbb C^3 \to \mathbb C^3\) and the \(f \circ x^{-1}\) part from \(\mathbb C^3 \to \mathbb C\). So it is consistent, and allows us to use the multidimensional chain rule:

\[=\left.{\partial_m\left( f \circ x^{-1}\right)}\right|_{\left(x \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1})(x\left(\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)\right)} \; \cdot \; \partial_i \left.{\left( x^m \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right)}\right|_{x\left(\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)}\]

where \(\partial_i \left( x^m \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right)\) is the inner derivative, and \(\partial_m\left( f \circ x^{-1}\right)\) the outer derivative, which is evaluated at \(\left( x \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right)(x\left(\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)) = (a,b,c).\)

Now going one step further, and considering that the expression takes a point in the chart \((e,f,g),\)

\[\left( x^m \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right)\, (e,f,g) = x^m \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \begin{bmatrix}e & f \\ g & \frac{1 + fg}{e}\end{bmatrix} = x^m \circ \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}e & f \\ g & \frac{1 + fg}{e}\end{bmatrix} = x^m \circ \begin{bmatrix}ae + bg & af + \frac{b + (1 + fg)}{e}\\ ce + dg & cf + \frac{d(1 + fg)}{e}\end{bmatrix}\]

and the \(m\)-th coordinate is

\[=\left( ae + bg , af + \frac{b(1 + fg)}{e} , ce + dg \right)^m\]

and therefore at \(m = 1,\) we are at \(ae + bg ,\) and deriving first to \(e,\) then to \(f\) and finally to \(g\) we get \((a, 0 , b).\) Next we do the same for \(m = 2\), which means \(af + \frac{b(1 + fg)}{e}\), etc.

We see that the expression

\[\partial_i \left.{\left( x^m \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right)}\right|_{x\left(\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)}= \partial_i \left.{\left( x^m \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right)}\right|_{(1,0,0)}= \left.{\begin{bmatrix} a & 0 & b \\ \frac{-b(1 + fg)}{e^2} & a + \frac{bg}{e} & \frac{bf}{e} \\ c & 0 & d\end{bmatrix}}\right|^m_{i\phantom{m}(1,0,0)}=\left.{\begin{bmatrix} a & 0 & b \\ -b & a & 0 \\ c & 0 & d\end{bmatrix}}\right|^m_{\phantom{m}i}\]

we get

\[\small \left [l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}*} \left(\frac{ \partial }{\partial x^i} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right ]_{\tiny \begin{bmatrix}a & b \\ c & d\end{bmatrix}}\; f = \left.{\partial_m\left( f \circ x^{-1}\right)}\right|_{\left(x \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1})(x\left(\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)\right)} \; \cdot \; \partial_i \left.{\left( x^m \circ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \circ x^{-1}\right)}\right|_{x\left(\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)}= \left(\frac{\partial}{\partial x^m}\right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}}f \; \cdot \; \left.{\begin{bmatrix} a & 0 & b \\ -b & a & 0 \\ c & 0 & \frac{1 + bc}{a}\end{bmatrix}}\right|^m_{\phantom{m}i}\]

with \(m\) being rows, and \(i\) columns.

Therefore for the first coordinate-induced basis tangent vector at the identity:

\[l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}*} \left(\frac{ \partial }{\partial x^1} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}} = a \left(\frac{\partial}{\partial x^1} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} - b \left(\frac{\partial}{\partial x^2} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} + c \left(\frac{\partial}{\partial x^3} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}}\]

For the second coordinate:

\[l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}*} \left(\frac{ \partial }{\partial x^2} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}} = a \left(\frac{\partial}{\partial x^2} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \]

and the third:

\[l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}*} \left(\frac{ \partial }{\partial x^3} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}} = b \left(\frac{\partial}{\partial x^1} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} + d \left(\frac{\partial}{\partial x^3} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}}\]

Therefore we constructed a vector field from a single vector at the identity.

Now we can calculate the bracket

\[\left [ \left(\frac{\partial}{\partial x^1}\right)_{\tiny \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}, \left(\frac{\partial}{\partial x^2}\right)_{\tiny \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right ] f:= l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}^{-1}*} \left [ l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}*} \left(\frac{ \partial }{\partial x^1} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}} , l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}*} \left(\frac{ \partial }{\partial x^2} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right ] f\]

and because the expressions on the LHS are vector fields,

\[\small=l_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}^{-1}*} \left[ a \left(\frac{\partial}{\partial x^1} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} - b \left(\frac{\partial}{\partial x^2} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} + c \left(\frac{\partial}{\partial x^3} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \; , \; a \left(\frac{\partial}{\partial x^2} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \right]f \\\\= \left.{\left[ a \left(\frac{\partial}{\partial x^1} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} - b \left(\frac{\partial}{\partial x^2} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} + c \left(\frac{\partial}{\partial x^3} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \; , \; a \left(\frac{\partial}{\partial x^2} \right)_{\tiny\begin{bmatrix}a & b \\ c & d\end{bmatrix}} \right]}\right|_{id}f\]

which can be worked out, keeping in mind that

\[\left(\frac{\partial}{\partial x^1}\right)_{\tiny \begin{bmatrix}a & b \\ c & d \end{bmatrix}}f=\partial_i(f \circ x^{-1})_{(a,b,c)}\]

The final calculations involve,

\[\left[ \left(\frac{\partial}{\partial a} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}, \left(\frac{\partial}{\partial b} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right]f = \left.{\left[ a \frac{\partial}{\partial a} - b \frac{\partial}{\partial b} + c \frac{\partial}{\partial c}, a \frac{\partial}{\partial b}\right]}\right|_{id} f\]

\[\left[ \left(\frac{\partial}{\partial a} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}, \left(\frac{\partial}{\partial c} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right]f = \left.{\left[ a \frac{\partial}{\partial a} - b \frac{\partial}{\partial b} + c \frac{\partial}{\partial c}, b \frac{\partial}{\partial a} + \frac{1 + bc}{a} \frac{\partial}{\partial c}\right]}\right|_{id} f\]

\[\left[ \left(\frac{\partial}{\partial b} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}, \left(\frac{\partial}{\partial c} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right]f = \left.{\left[ a \frac{\partial}{\partial b} , b \frac{\partial}{\partial a} + \frac{1 + bc}{a} \frac{\partial}{\partial c}\right]}\right|_{id} f\]

Evaluating the last expression,

\[=a \frac{\partial}{\partial b} \left(b \frac{\partial}{\partial a} f + \frac{1 + bc}{a} \frac{\partial}{\partial c} f\right) - \left(b \frac{\partial}{\partial a} + \frac{1 + bc}{a} \frac{\partial}{\partial c}\right)\left(a \frac{\partial}{\partial b}f \right)\]

further

\[=a \frac{\partial}{\partial a}f+ ab \frac{\partial^2}{\partial b \partial a}f + c \frac{\partial}{\partial c} f + a \frac{1 + bc}{a}\frac{\partial^2}{\partial b \partial c}f - b \frac{\partial}{\partial b}f - ba \frac{\partial^2}{\partial a \partial b}f - (1 + bc) \frac{\partial^2}{\partial c \partial b}f\]

which simplifies to

\[\left[ \left(\frac{\partial}{\partial b} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}, \left(\frac{\partial}{\partial c} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right]f =\left.{\left( a \frac{\partial}{ \partial a} - b \frac{\partial}{\partial b} + c \frac{\partial}{\partial c} \right)}\right|_{id} f=\left.{\left( a \frac{\partial}{ \partial a} \right)}\right|_{id} f\]

Similarly,

\[\left[ \left(\frac{\partial}{\partial a} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}, \left(\frac{\partial}{\partial b} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right]f = 2 \left( \frac{\partial}{\partial b}\right)_{id} f\]

and

\[\left[ \left(\frac{\partial}{\partial a} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}, \left(\frac{\partial}{\partial c} \right)_{\tiny\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}\right]f =\left( -b \frac{\partial}{\partial a} - \frac{1 + bc}{a} \frac{\partial}{\partial c} \right)_{id} f = - 2 \left( \frac{\partial}{\partial c} \right)_{id} f\]

In summary,

\[[X_1, X_2] = 2 X_2\] \[[X_1, X_3] = -2 X_3\] \[[X_2, X_3] = X_1\]

where \(X_i = \left( \frac{\partial}{\partial x^i} \right)_{id}\) is the \(i\)-th abstract differential coordinates tangent vector at the identity.

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NOTE: These are tentative notes on different topics for personal use - expect mistakes and misunderstandings.