# Volume form on a sphere.

Let $S^n(r)$ be the sphere of radius $r$ , $x_1^2 + ... + x_{n+1}^2 = r^2$ and let $$w = \frac{1}{r} \sum_{i=1}^{n+1} (-1)^{i-1} x_i dx_1 \cdots\hat{dx_i},\cdots dx_{n+1}$$

Write $S^n$ for the unit sphere $S^n(1)$. Compute the integral $$\int\limits_{S^n}w$$ and conclude that $w$ is not exact.

For me this is a hard exercise. I understand that from (a) we obtain an explicit formula for the generator of the top cohomology of $S^n$ (although not as a bump form).

• Are you familiar with Stoke's theorem for differential forms? – Eric O. Korman Nov 5 '13 at 1:24
• yes, how can i use it? – Rachel Nov 5 '13 at 2:26
• @Knight: Try to calculate $dw$. – user99914 Nov 5 '13 at 2:36
• Someone can give more details and i can not start , i tried compute dw but really i stuck in this problem – Knight 48 mins ago – Rachel Nov 5 '13 at 3:54

John Ma interpreted the form $w$ differently than I believe was intended. He interpreted $w$ as a constant called $1/r$ times an $n$-form. He then ignored the constant and correctly used Stokes theorem to find the integral of said $n$-form over the sphere. I believe that, in fact, $r$ was intended to be a function of the $x$ variables. In any case, let me add here a computation of $dw$ with this latter interpretation of $w$.

If $w$ depends on $r$ then one gets an additional term in John Ma's formula for $dw$. Indeed, since

$$r^{-1} = |x|^{-1}= \left(\sum_j x_j^2 \right)^{-1/2}$$

we find that

$$d r^{-1} = -|x|^{-3} \cdot \sum_j x_j dx_j$$

Thus, the exterior derivative $dw$ has the term indicated in the answer of John Ma plus

$$-|x|^{-3} \left(\sum x_j dx_j \right) \wedge \left(\sum_i (-1)^{i-1} x_i \alpha_i \right)$$

where $\alpha_i$ denotes the wedge product of all of the $dx_j$'s except for $dx_i$. Since $dx_i \wedge dx_{i+1} = -dx_{i+1} \wedge dx_i$ and $dx_i \wedge dx_i=0$, the last displayed expression simplifies to

$$-|x|^{-3} \sum_{i,j} (-1)^{i-1} x_jx_i~ dx_j \wedge \alpha_i~ =~ -|x|^{-3}\sum_j x_j^2~dV~ =~ -|x|^{-1} dV$$

where $dV$ is the product of all $dx_i$'s without exception.
So we have

$$dw~ =~ \frac{n}{r} \cdot dV.$$

Because $w$ depends on $r$, then applying Stokes theorem

$$\int_{\Omega} d \omega = \int_{\partial \Omega} \omega$$

requires some care. Indeed, $w$ is not defined---as is---on the closed unit ball $B$. In particular, it is not defined at $x=0$. If $n>0$, then by setting $\omega_x \equiv 0$ for $x=0$ we obtain a continuous extension to ${\mathbb R}^{n+1}$. However, this extension is not differentiable if $n=1$. For $n>0$, one can apply Stokes theorem on the domain $B_1-B_{\epsilon}$ where $B_r$ denotes the ball of radius $r$. Then Stokes theorem reads

$$\int_{B_1-B_{\epsilon}} dw~ = \int_{\partial B_1} w~ +~ \int_{\partial B_{\epsilon}} w$$

where one should pay attention to orientations of the boundary. For $n>0$, $\|\omega_x\|$ tends to zero as $x$ tends to zero, and hence the latter integral tends to zero. Thus we have

$$\int_{B_1} \frac{n}{r} \cdot dV~ =~ \int_{\partial B_1} w$$

Also if $w$ depends on $r$, then we can use polar coordinates to evaluate the integral. That is, write $dV= r^n dr \wedge d \omega$ where $d \omega$ is the usual measure on the unit sphere. The above integral becomes

$$\int_{\partial B_1} \int_{0}^1 n \cdot r^{n-1} \cdot dr\, d\omega~ = \int_{\partial B_1} d \omega~ =~ {\rm Vol} (\partial B_1)$$

If $w$ were exact, then $w=d \alpha$ and by Stokes

$$\int_{\partial B_1} w~ =~ \int_{ \partial B_1} d \alpha =\int_{\partial (\partial B_1)} \alpha$$

But the sphere has no boundary and so the latter integral is zero. This contradicts the fact that the sphere has nonzero volume.

Of course, if you're only interested in showing that the form is not exact. then it's easier not to consider $w$ but rather $r \cdot w$ as John Ma does.

• Well, all three objections are just really one (except for $n=0$ case which I did not think of) - that I treat $r$ as a constant while you do not. To me, since it says "Let $\mathbb S^n(r)$ be the sphere of radius $r$, I assume that $r$ is just a given number. So if $r=2$ for example, $w = \frac{1}{2} \sum_{i=1}^{n+1} (-1)^{i-1} x_i dx_1 \cdots\hat{dx_i},\cdots dx_{n+1}$. – user99914 Nov 26 '16 at 8:55
• @ John Ma. My main objection is that you can not treat $r$ as a constant if you hope to apply Stokes theorem by taking $dw$. Because you treat $r$ as a constant, you end up with the incorrect formula $dw=r^{-1} (n+1)dV$. The correct formula is $dw= r^{-1} n dV$. The case $n=1$ is also problematic because $w$ is not diffferentiable at $n=0$. – Chris Judge Nov 26 '16 at 12:20
• $\omega$ is a smooth $n$ form on $\mathbb R^{n+1}$ (if $r$ is a constant). So $d\omega$ is also smooth on $\mathbb R^{n+1}$ and one can apply Stokes theorem. – user99914 Nov 26 '16 at 12:49
• In the original question $w$ was defined with $r$ on the right hand side. If you want to apply Stoke theorem to the ball of radius $r$, then you the first displayed formula in your answer $r$ is not a constant. – Chris Judge Nov 26 '16 at 18:57
• The thing to do is to define another 1-form say $\omega$ by $\omega= r w$. Now $\omega$ is smooth and $d \omega$ can be calculated as you like. This is probably what you wanted to do, but it's not clear. The question was about $S(1)$ and so it seems clear that the questioner was thinking of $r$ as being variable. – Chris Judge Nov 26 '16 at 19:03

We have
$$w = \frac{1}{r} \sum_{i=1}^{n+1} (-1)^{i-1} x_i dx_1 \cdots\hat{dx_i},\cdots dx_{n+1} \ ,$$

thus

$$dw = \frac{1}{r} \sum_{i=1}^{n+1} (-1)^{i-1} dx_i dx_1 \cdots\hat{dx_i},\cdots dx_{n+1} = \frac{1}{r} \sum_{i=1}^{n+1} dx_1\cdots dx_{n+1} = \frac{n+1}{r} dx_1 \cdots dx_{n+1}\ .$$

Note that the second equality holds because in commuting those $dx_j$ we have the rule

$$dx_i dx_j = - dx_j dx_i\ .$$

Now by Stokes theorem, as $\mathbb S^n = \partial B^{n+1}$,

$$\int_{\mathbb S^n} w = \int_{\partial B^{n+1}} w = \int_{\mathbb B^{n+1}} dw = (n+1) V(B^{n+1})\neq 0$$

Thus $w$ when restricted to $\mathbb S^n$ is not exact (that it is not exact in $\mathbb R^{n+1}$ is obvious as $dw \neq 0$). The reason is again by Stokes theorem: if $w = d\alpha$, then

$$\int_{\mathbb S^n} w = \int_{\mathbb S^n} d\alpha = 0\ .$$

But we have seen that this is not zero.

• Doesn't the $1/r$ in the definition of $w$ have to be differentiated also? – Muphrid Nov 5 '13 at 4:44
• I think $r$ is fixed (See the first line of the question) – user99914 Nov 5 '13 at 4:47