First part of the proof that $F^d\beta=dF^\beta$

Question

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Where has the $dy^j$ gone in the highlighted equation?

I would have thought the highlighted equation should be $\displaystyle (F^*dg)(x) = \frac{\partial F^j}{\partial x^i}(x)\frac{\partial g}{\partial y^j}(F(x))dy^j dx^i$

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The formula at the bottom of your post kind of answers this. I'm not sure why you think $dy^j$ should stay. Notice that you have too many $j$ indexed variables for Einstein summation. The $\beta_j$ are $\frac{\partial g}{\partial y^j}$ here — Callus - Reinstate Monica, Nov 10 '14 at 11:28
take a look at http://math.stackexchange.com/questions/456586/fdx-i-sum-j-1l-frac-partial-f-i-partial-y-j-dy-j-df-i — janmarqz, Nov 10 '14 at 20:02

score 1 · Accepted Answer · answered Nov 10 '14 at 19:46

1

Hint. For a one-form $P_sdx^s$ we have the two-form $$dP=\frac{\partial P_s}{\partial y^t}dy^t\wedge dx^s.$$ Note that there are two sum indexes.

answered Nov 10 '14 at 19:46

janmarqz

10,538

First part of the proof that $F^*d\beta=dF^*\beta$

1 Answers1

First part of the proof that $F^d\beta=dF^\beta$