Where has the $dy^j$ gone in the highlighted equation?
I would have thought the highlighted equation should be $\displaystyle (F^*dg)(x) = \frac{\partial F^j}{\partial x^i}(x)\frac{\partial g}{\partial y^j}(F(x))dy^j dx^i$
2026-04-04 13:21:28.1775308888
First part of the proof that $F^*d\beta=dF^*\beta$
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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.