In local coordinates, for
$$
\alpha=\frac1{r!}\alpha_{i_1\ldots i_r}\,
dx^{i_1}\wedge\cdots\wedge dx^{i_r},
$$
the <exterior derivative> is
$$
d\alpha=\frac1{r!}\frac{\partial\alpha_{i_1\ldots i_r}}{\partial x^j}
dx^j\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_r}.
$$
Applying $d$ again gives symmetric second partial derivatives contracted with the antisymmetric wedge $dx^k\wedge dx^j$, hence $d^2=0$. Expanding coefficients and moving $d\alpha$ past a degree-$r$ form gives the graded Leibniz rule
$$
d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^r\alpha\wedge d\beta.
$$

For a function $f$ and smooth $F:X\to Y$, the chain rule gives
$$
d(F^*f)=d(f\circ F)=F^*(df).
$$
Every differential form is locally a sum of products $f_0\,df_1\wedge\cdots\wedge df_r$. Since pullback preserves products and wedge products, the function case and the graded Leibniz rule imply
$$
d(F^*\alpha)=F^*(d\alpha)
$$
for every form.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-115/2/a/solution