jump to navigation

Symmetric connections, corrected version November 9, 2009

Posted by Akhil Mathew in differential geometry, MaBloWriMo.
Tags: , , ,
trackback

My post yesterday on the torsion tensor and symmetry had a serious error.  For some reason I thought that connections can be pulled back.  I am correcting the latter part of that post (where I used that erroneous claim) here. I decided not to repeat the (as far as I know) correct earlier part.

Proposition 1 Let {s} be a surface in {M}, and let {\nabla} be a symmetric connection on {M}. Then\displaystyle \frac{D}{\partial x} \frac{\partial}{\partial y} s = \frac{D}{\partial y} \frac{\partial}{\partial x} s.\ \ \ \ \ (1) 

 

Assume first {s} is an immersion. Then at some fixed {p \in U}

\displaystyle \frac{D}{\partial x} \frac{\partial}{\partial y} s = (\nabla_{X} Y) \circ s,

 if {X} is a vector field on some neighborhood of {s(p)} which is {s}-related to {\frac{\partial}{\partial x}}, and {Y} similarly for {\frac{\partial}{\partial y}}. Similarly,

\displaystyle \frac{D}{\partial y} \frac{\partial}{\partial x} s = (\nabla_{Y} X) \circ s.

 The difference between these two quantities is {T(X,Y) \circ s=0}, because [X,Y] \circ s = 0 by a general theorem about f-relatedness preserving the Lie bracket for f a morphism.   As before, we can approximate s by an immersion at p, and we get the general case.

That was much easier than what I was trying to do earlier. Blogging is a learning experience.

Advertisements

Comments»

No comments yet — be the first.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: