Integration of one-form

Question

I am trying to compute

$$\int_C i^*\eta$$

$\eta=(x^2+y^2)dz$ and $C=\{(x,y,-1): x^2+y^2=1\}$ and $i$ is the inclusion map

This is what I did $$\int_{-1}^{-1}(x^2+y^2) dz=0 $$ Is this correct?

Answer 1

The circle $C=\{(x,y,-1)\colon x^2+y^2=1\}$ admits a parametrization by $(x,y,z)=(\cos t,\sin t,-1),$ for $0\leq t\leq 2\pi.$ Therefore if $$\eta=(x^2+y^2)\,dz$$ we compute $$i^*\eta=(\cos^2t+\sin^2t)\,d(-1)=0.$$

Hence $$\int_0^{2\pi} i^*\eta = 0.$$

Intuitively, a 1-form with only a $dz$ component is a 1-form which measures only $z$-components of directional derivatives. So it vanishes along any curve with constant $z$-component.

Answer 2

If $c:I \to C$ is given by $c(t)=(\cos 2\pi t,\sin 2\pi t, -1)$ where $I=[0,1]$.

Then your integral is $\int_I (i \circ c)^*\eta$

For the vector $\mathbf{v}$ in the 1-D tangent space of I, $$(i \circ c)^* \eta(\mathbf{v})=\eta((i \circ c)_*(\mathbf{v})=(x^2+y^2)dz( -2 \pi v \sin 2 \pi t, 2 \pi v \cos 2\pi t, 0)=0$$