4 ms·
In Lie theory, given a Lie group, there's a general notion of an "exponential function", which maps elements in the tangent space to the identity to "full" elem
by nilkn 10y ago
In Lie theory, given a Lie group, there's a general notion of an "exponential function", which maps elements in the tangent space to the identity to "full" elements of the group.
In this case, our group is the unit circle in the complex plane. This is not circular logic, by the way. If a, b are complex numbers on this circle, then |a| = |b| = 1 and so |ab| = |a||b| = 1.
The identity of the group is the complex number 1 (with plane coordinates (1,0)). So the tangent space to the identity is the vertical line 1 + it, for t a real-valued parameter. In two-dimensional coordinates, that expression looks like (1,0) + t * (0,1).
(To be completely precise, the tangent space is a vector space, not a line displaced from the origin. In particular, the tangent space must contain a zero vector.)
If v is an element of this tangent space, then in Lie theory the exponential of v, exp(v), is defined to be g(1), where g is the unique geodesic on the circle (passing through the identity element 1) whose velocity at time 0 is v.
Visually, this is sort of like taking the tangent vector v = ti, placing its base at 1, then wrapping it around the circle and marking where its endpoint ends up at. If the vector is v = pi*i, then it has length pi, so it will end up demarcating an arc length of pi on the circle. Since we're working with the unit circle, this takes us straight to (-1,0).
I'm still leaving a lot out, of course -- most importantly why this notion of exponential has anything to do with the ordinary one.
- chombier 10y agoI was about to write a similar story, but then I realized you need a Riemannian metric (and exponential) for everything arclength-related. IIRC since the unit circle is a compact Lie group, there is a bi-invariant Riemannian metric whose exponential is the Lie group exponential and we land back on our feet, but the Lie group structure alone is not sufficient.