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- Newsgroups: sci.math
- Path: sparky!uunet!enterpoop.mit.edu!galois!riesz!jbaez
- From: jbaez@riesz.mit.edu (John C. Baez)
- Subject: Re: Tensor derivatives
- Message-ID: <1992Dec21.181334.4239@galois.mit.edu>
- Sender: news@galois.mit.edu
- Nntp-Posting-Host: riesz
- Organization: MIT Department of Mathematics, Cambridge, MA
- References: <9235215.10767@mulga.cs.mu.OZ.AU>
- Date: Mon, 21 Dec 92 18:13:34 GMT
- Lines: 26
-
- In article <9235215.10767@mulga.cs.mu.OZ.AU> psm@burra.ee.mu.OZ.AU (Phil Malin) writes:
- >Hi all.
- >
- >This question may not make sense, but anyway...
- >
- >I've read about covariant and contravariant derivatives. Does it
- >make sense to have mixed tensor derivatives? i.e. taking the derivative
- >of a function w.r.t. a type (1,1) tensor, etc... if the function is
- >a function of this tensor? Does this question make sense?
-
- >If, on the other hand, this idea shows a total lack of understanding
- >of the subject please inform me gently (no flames :-).
-
- I have heard about covariant and contravariant tensors, and covariant
- derivatives, but not contravariant derivatives. So this post
- shows a total lack of understanding of the subject -- either on Malin's
- part or my own, but I'm not sure which. :-) What would a contravariant
- derivative be, anyway? Ordinarily one takes a derivative with respect
- to a tangent vector, so would this amount to taking the derivative with
- respect to a cotangent vector? Most odd. Of course if you have a
- Riemann metric around you can convert tangent vectors into cotangent
- vectors and vice versa ad libitum, so that *could* give "contravariant
- derivative" a meaning.
-
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-
-