## Determinants and traces – 1

I wonder how many posts I’ll label “Subject matter – 1” and how many sequels I’ll end up actually writing.  Last time, I wrote that I had started thinking about the trace of a linear map.  I finally found a few moments, after the kiddo finally went to sleep, to play with an example.

Consider the Lie group of 2×2 invertible matrices with real entries, $GL(2,\mathbb{R})$.  The determinant is a differentiable homomorphism from GL to R.  As such, it’s differential is a map between tangent spaces of GL and R.  Let’s look at a rather particular tangent space, the Lie algebras $\mathfrak{gl}$ and $\mathbb{R}$.  Let $\vec{v}\in\mathfrak{gl}$.  It’s 4-dimensional (just like $GL_2$).  Let’s use the following coordinates around the identity element of GL,

$\begin{pmatrix} a &b \\ c& d \end{pmatrix}$

To find a basis for our Lie algebra in terms of the coordinates around the identity we take such a matrix as above and differentiate in each direction and evaluate at I.  As an example consider the ‘a’ direction,

Each direction will result in a basis vector for $\mathfrak{gl}$,

So, given arbitrary coefficients, an element of $\mathfrak{gl}$ would look like,

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

But this time with no constraint on the determinant.  $\mathfrak{gl}\cong M(n,\mathbb{R})$.  I don’t want to break the flow of this post but I’m going to do a Lie groups and Lie algebras -3 soon where I address some of my own questions from before.  Notice, for the moment, how easy it is to obtain the above elements of $\mathfrak{gl}$ but that this vector space is isomorphic to $\mathbb{R}^4$ and that doesn’t give us diddly.  This is why we need some sort of product, otherwise there’s nothing special about any of the tangent spaces of GL.

So our typical tangent vector, using local coordinates, would look like,

$\vec{v}=v_1\frac{\partial}{\partial a}+v_2\frac{\partial}{\partial b}+v_3\frac{\partial}{\partial c}+v_4\frac{\partial}{\partial d}$

While our typical tangent vector in the Lie algebra of R would look like $\vec{w}=w\frac{d}{dt}$ for coordinate t.  Let $g:\mathbb{R}\to\mathbb{R}$.  Denote the differential of the determinant as det*.  So $det^*(\vec{v})=\vec{w}$  The image, $\vec{w}$ acts on functions of R as $\vec{w}(g)=w\dot{g}(0)$ where the dot denotes differentiation with respect to t. Let’s figure out what w should be given an arbitrary $\vec{v}$ in the Lie algebra of GL.  We’ll use the above mentioned function, g, as dummy input.

$\vec{w}(g)=w\dot{g}=(det^*\vec{v})(g)=\vec{v}(g\circ det)$

The important thing to remember in the next step is that when a tangent vector acts on a function that it’s not just differentiation, but differentiation and evaluation.  The tangent vectors in the Lie algebra differentiate functions on the Lie group and then evaluate at the identity element, which for us is when a=1, b=0, c=0, and d=1.

$\vec{v}(g\circ det)=\vec{v}(g(ad-bc)=1\cdot\dot{g}v_1-0\cdot\dot{g}v_2-0\cdot\dot{g}v_3+1\cdot\dot{g}v_4$

Which reduces to $\dot{g}(v_1+v_4)$  This is just the trace of

$\begin{bmatrix} v_1 & v_2\\ v_3 & v_4 \end{bmatrix}$

So one way of looking at the trace is as the differential of the determinant evaluated at the identity.  So it’s  a local linearization of the deteminant for values near the identity.  The determinant of the identity is 1, for any of these matrix groups.  But the trace for the Lie algebras is going to vary.  As a spoiler, the Lie algebra $\mathfrak{sl}$ contains only matrices of trace zero.  So there’s more.  The determinant of a matrix also tells us how it scales the space its acting on and whether it acts as a reflection or not.  Does the trace give us any information concerning this?  We’ll see.