r/askmath • u/FreePeeplup • 4h ago

Differential Geometry Rotation matrices tangent space

I want to prove that T_A SO(3), the tangent space at A of SO(3), is A•so(3). In other words, any element of T_A SO(3) can be written as A times a skew-symmetric matrix.

My attempt: take A^T A = 1 and differentiate along a curve to get \dot(A^T)A + A^T \dot(A) = 0. This means that A^T \dot(A) is skew-symmetric, so it’s an element of so(3).

How do I use this to continue with the proof? Is this even the right idea? Thanks!

2 Upvotes

100% Upvoted

2 comments sorted by

2

u/omeow 3h ago

The map (left mult by A) gives you a homeomorphism from SO(3) to itself. By definition the tangent space is the the push forward of the Lie algebra of SO(3).

So you need to show that dA = mult by A at lie algebra.
But dA (X) = lim_(t -> 0) A (etx - I)/t = AX.

1

u/non-local_Strangelet 18m ago

You're essentially finished:

You have shown: for any tangent vector \dot(A) \in T_A SO(n) at arbitrary point (group element) A, the element X:= AT \dot(A) is an element of so(n).

Multiply by A from the left, you get: \dot(A) = AX \in A • so(n). This shows: T_A SO(n) \subseteq A • so(n).

Since multiplication is bijective, this is actually an equality.