Tangent vector to a curve (Differential geometry/Lie theory).

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SUMMARY

The discussion focuses on finding the tangent vector to the curve defined by the matrix function c(s) in the special orthogonal group SO(3) at the identity matrix I_3. The tangent vector X is defined using the differential operator, specifically X[f] = df(c(s))/ds|_{s=0}. The participants clarify that the correct approach involves computing the matrix derivative dc/ds at s=0, leading to the conclusion that the corresponding tangent vector is represented as -∂/∂x^{12} + ∂/∂x^{21}, which is an element of the Lie algebra so(3) associated with T_eSO(3).

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  • Understanding of Lie groups and Lie algebras, specifically in the context of SO(3).
  • Familiarity with differential operators and their application in differential geometry.
  • Knowledge of matrix calculus and partial differentiation.
  • Basic concepts of tangent vectors in the context of differentiable manifolds.
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  • Study the properties of Lie algebras, particularly so(3), and their relationship with Lie groups.
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  • Explore matrix calculus techniques, focusing on derivatives of matrix-valued functions.
  • Investigate the geometric interpretation of tangent vectors in differential geometry.
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B L
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Homework Statement



Let c(s) = \left( \begin{array}{ccc}<br /> \cos(s) &amp; -\sin(s) &amp; 0 \\<br /> \sin(s) &amp; \cos(s) &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \end{array} \right) be a curve in SO(3). Find the tangent vector to this curve at I_3.

Homework Equations



Presumably, the definition of a tangent vector as a differential operator would be useful here:
If X is the tangent vector to c(s) at I_3 (i.e. s = 0), then for functions f: SO(3) \rightarrow \mathbb{R},
X\left[ f \right] = \frac{df\left(c\left(s\right)\right)}{ds}|_{s=0}

The Attempt at a Solution


The fact that this problem is found in the section of the textbook (Nakahara, Geometry, Topology and Physics) dealing with Lie groups and Lie algebras, together with the fact that the problem asks for the tangent vector to c(s) at the identity leads me to think that we are supposed to use the Lie algebra in some way. However, I can't see how one would do that, so I've tried proceeding naively from the definition:
X\left[ f \right] = \frac{df\left(c\left(s\right)\right)}{ds}|_{s=0}
= \frac{df}{dc} \frac{dc}{ds}|_{s=0}
= \frac{df}{dc} \left( \begin{array}{ccc}<br /> 0 &amp; -1 &amp; 0 \\<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 \end{array} \right)

I have no idea how to proceed from here.

Thanks for any and all help!

EDIT: I quickly realized how trivial this is.
 
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You are already done. Sort of. X[f] would indeed be df(c(s))/ds at s=0. But that would be a real number. Writing something like df/dc is a little misguided. That would be the derivative of a real function with respect to a matrix. It would be tough to clearly define that. I think all they really want is the matrix dc/ds at s=0. And that you already have.
 
Am I correct in thinking that the corresponding tangent vector (i.e. element of the Lie algebra so(3) \cong T_eSO(3)) would be -\frac{\partial}{\partial x^{12}} +\frac{\partial}{\partial x^{21}}?

Thanks for the help.
 
B L said:
Am I correct in thinking that the corresponding tangent vector (i.e. element of the Lie algebra so(3) \cong T_eSO(3)) would be -\frac{\partial}{\partial x^{12}} +\frac{\partial}{\partial x^{21}}?

Thanks for the help.

Yes, you could write it that way. That would tell you how X acts on a function f:SO(3)->R.
 
I know why I went wrong - what I didn't realize is that by differentiation wrt c, I really meant partial differentiation wrt c^ij (and differentiation of c^ij wrt s) As soon as you do that the problem is trivial.
 

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