Let $$T=\{(x,y,z,w,)\in R^4:x^2+y^2=z^2+w^2=\frac{1}{\sqrt 2}\}$$ and $$\omega=dx\land dy + dz\land dw$$ in $\mathbb R^4$. How do I find a two-dimensional chain $C$ where $T$ is its trace? And how can I calculate $\int_C \omega$?
I missed some lectures on this topic because I was preparing for another exam, some help would be much appreciated, thank you.
Computing the integral is easy. Pick the following natural parametrization of $T$:
$$(x, y, z, w) = \frac 1 {\sqrt[4] 2}(\cos s, \sin s, \cos t, \sin t) $$
with $s,t \in [0, 2 \pi)$.
Notice that
$$\begin{align} \sqrt[4] 2 \omega &= \Bbb d (\cos s) \wedge \Bbb d (\sin s) + \Bbb d (\cos t) \wedge \Bbb d (\sin t) \\ &= (- \sin s \ \Bbb d s) \wedge (\cos s \ \Bbb d s) + (- \sin t \ \Bbb d t) \wedge (\cos t \ \Bbb d t) \\ &= (-\sin s \cos s) \ (\Bbb d s \wedge \Bbb d s) + (-\sin t \cos t) \ (\Bbb d t \wedge \Bbb d t) \\ &= 0 \end{align}$$
so that $\omega = 0$ (because $\Bbb d s \wedge \Bbb d s = 0$ and $\Bbb d t \wedge \Bbb d t = 0$ and the value of a form in a point is independent of the coordinates chosen around that point), hence its integral is also $0$ (notice that you don't have to "integrate" anything, that's the whole idea of the problem: it looks like a problem about integration, but it turns out to be something else - much simpler, in fact).
In fact, this was to be expected: notice that $T = \{(z, z') \in \Bbb C \times \Bbb C \mid |z| = |z'| = 1\}$, so that $T = S^1 \times S^1$ where $S^1 = \{z \in \Bbb C \mid |z| = 1\}$ (the unit circle in the complex plane). The $\Bbb d x \wedge \Bbb d y$ part, being independent of $u$ and $v$, lives on the first factor $S^1$ of $T$. Since $S^1$ is a curve (so it has dimension $1$) and $\Bbb d x \wedge \Bbb d y$ is a $2$-form, and $2$-forms vanish on $1$-dimensional objects, it is clear now that $\omega = 0$, in spite of the fact that it "looks non-zero".