Is the kernel of the coderivative infinite-dimensional?


Let $M,N$ be smooth $n$-dimensional Riemannian manifolds (perhaps with smooth boundary), and let $\, f:M \to N$ be a smooth immersion. Let $\Omega^k(M,f^*TN)$ be the space of $f^*TN$-valued $k$-forms.

Let $d:\Omega^k(M,f^*TN) \to \Omega^{k+1}(M,f^*TN)$ be the covariant exterior derivative associated with the pullback connection of the Levi-Civita connection on $N$ (via $f$), and let $\delta$ be its adjoint.

Denote $\delta_{k}:\Omega^k(M,f^*TN) \to \Omega^{k-1}(M,f^*TN)$.

Question: Is $\ker \delta_1$ is infinite dimensional?

Since $\delta=\star d \star$ (up to sign), $\dim(\ker \delta_1)=\dim(\ker d_{n-1})$. So, this is a question about $d$.

Edit:

In the special case where $N$ is flat, $\delta_{k} \circ \delta_{k+1}=0$, so $\operatorname{Image}(\delta_{k+1}) \subseteq \ker(\delta_k)$.

In particulr, $\operatorname{Image}(\delta_{2}) \subseteq \ker(\delta_1)$.

Is $\operatorname{Image}(\delta_{2})$ is infinite-dimensional? Can we at least something when $N=\mathbb{R}^n$? or even when $M=N=\mathbb{R}^n$?

Let's see what happens when $M=N=\mathbb{R}^n,f=\operatorname{Id}$:

$$d_{n-1}(\sigma)(e_1,...,e_n)=\sum_{j=1}^{d} (-1)^{j+1} \nabla^{T\mathbb{R}^n}_{e_j} \big( \sigma(e_1,...,\hat{e_j},...,e_d) \big)$$

Write $\sigma=(-1)^{j+1}a_j^idx^1 \wedge \dots \wedge \hat{dx^j} \wedge \dots \wedge dx^d \otimes e_i, a_j^i \in C^{\infty}(\mathbb{R}^n)$.

Then $\sigma(e_1,...,\hat{e_j},...,e_d)=(-1)^{j+1}a_j^i e_i$, so

$$\nabla^{T\mathbb{R}^n}_{e_j} \big( \sigma(e_1,...,\hat{e_j},...,e_d) \big)= \nabla^{T\mathbb{R}^n}_{e_j}((-1)^{j+1}a_j^i e_i)=(-1)^{j+1}\frac{\partial a_j^i}{\partial x_j}e_i.$$

Thus, $$\sigma \in \ker d_{n-1} \iff \frac{\partial a_j^i}{\partial x_j}e_i=0$$ (in the last term there is a double summation, on $i,j$).

Since $e_i$ are independent, this is equivalent to $$\sum_{j=1}^d \frac{\partial a_j^i}{\partial x_j}=0 \, \text{ for all } \, i=1,...,d$$

Denoting $\bar a^i=(a_1^i,...,a_d^i)$, we get that $\operatorname{div} (\bar a^i)=0$.

I guess it shouldn't be too hard to see now that $\dim(\ker d_{n-1}) = \infty$. (By taking the $a_j^i$ to be constants, one immediately gets $\dim(\ker d_{n-1}) \ge n^2$. Since the condition on the divergence do not touch $\frac{\partial a_j^i}{\partial x_k}$ for $k \neq j$, it seems plausible that the dimension is indeed infinite. Perhaps someone can come with a slick argument to show this.)

The next case we should try is $M=N=\mathbb{R}^d$, and $f$ an arbitrary mapping....

Comment:

I know that $\operatorname{Cof}(df) \in \ker \delta_1$, where $\operatorname{Cof}(df)$ is the corresponding cofactor map of $df$: $$\Cof df= (-1)^{d-1} \star_{f^*TN}^{n-1} (\wedge^{n-1} df) \star_{TM}^1.$$

Since $f$ is an immersion, $\operatorname{Cof}(df) \neq 0$, so $\dim (\ker \delta_1 ) \ge 1$.

For my purposes, It would suffice to know that $\ker \delta_1$ always contains elements which are linearly independent of $\Cof df$.

• Myself I suspect that $\dim\ker(\delta)=\infty$ since $\dim \ker(d)\cap\ker(\delta)<\infty$ and $\dim\ker(d)$ tends to be infinite dimensional. – Liviu Nicolaescu Nov 20 '16 at 17:13

2 Answers

Suppose that the vector field $v$ is a non-vanishing covariantly constant on $N$, then its pullback $f^*v$ is a constant section of $f^*TN \to M$ with respect to the pulled back connection. Then, for any $(n-1)$-form $\alpha$ and $(n-2)$-form $\beta$ on $M$, we have $d_{n-2} (\beta \otimes f^*v) = (d\beta) \otimes f^*v$ and $d_{n-1} (\alpha \otimes f^*v) = (d\alpha) \otimes f^*v$, where I used $d$ to denote the usual de Rham exterior derivative. Then clearly, $\mathrm{im}(d) \otimes f^*v \subseteq \ker d_{n-1}$ is an infinite dimensional subspace.

On the other hand, suppose that there are no non-vanishing constant sections of $f^*TN \to M$. We might still be able to find some differential operator $q \colon \Gamma(E) \to \Gamma(\Lambda^{n-1}T^*M \otimes_M f^*TN)$ whose image lands inside $\ker d_{n-1}$, that is $d_{n-1} \circ q = 0$. The trick is to first dualize. As you noted, $d_{n-1}^* = d_0$, up to conjugation by invertible operators. Then any "compatibility operator" $q^*$, such that $q^* \circ d_0 = 0$, gives us a candidate $q = (q^*)^*$, because $d_{n-1} \circ q = d_0^* \circ (q^*)^* = (q^* \circ d_0)^*$.

The non-existence of non-trivial constant sections of $f^*TN \to M$ means that there exist integrability conditions for the equation $d_0 v = 0$. In particular, after some massaging, we can come up with an operator $C\colon \Gamma(T^*M \otimes_M f^*TN) \to \Gamma(f^*TN)$ such that $C\circ d_0 = \mathrm{id}$. As a consequence, we have the identity $(\mathrm{id} - d_0 \circ C) \circ d_0 = d_0 - d_0 \circ (C\circ d_0) = 0$. Hence we have found a compatibility operator $q^* = \mathrm{id} - d_0 \circ C$, from which $q = \mathrm{id} - C^* \circ d_{n-1}$. In conclusion, we have $\mathrm{im}(q) \subseteq \ker d_{n-1}$.

It remains to check that $\mathrm{im}(q)$ is indeed infinite dimensional. I don't have a slick argument for that, at the moment. But I'm sure that in individual cases it should not be hard to establish, at the very least by looking at the effect of $q$ on jets of higher and higher order at some fixed point $x\in M$.

• Thanks. Your solution is very interesting. I have some questions: (1) Is it known that $\operatorname{image}(d_{n-2})$ is infinite-dimensional? (I only know that $\operatorname{image}(d_{n-2})+\operatorname{image}(\delta_{n})$ is infinite-dim: This follows from the Hodge decomposition theorem together with the facts that the space of harmonic forms is finite-dim and the space of all forms of a certain degree is infinite-dim ). (2) I would really like to know more about what exactly do you mean by "integrability conditions", and how you lift them to build an operator $C$. – Asaf Shachar Apr 1 '17 at 6:40
• (Finally, I think there is a typo: you meant to write $d_1$ instead of $d_0$...) – Asaf Shachar Apr 1 '17 at 6:40
• I meant the operator $d_0\colon \Gamma(f^*TN) \to \Gamma(T^*M\otimes_M f^*TN)$. The identity $\beta \wedge (d_{n-1} \alpha) + (d_0\beta) \wedge \alpha = d_{n-1} (\beta \wedge \alpha)$ shows that, up to signs, $d_{n-1}^* = d_0$ (integrate both sides with compactly supported arguments to check). These formulas are for usual forms, but using appropriate covariant derivatives they work also for vector valued forms. – Igor Khavkine Apr 1 '17 at 9:07
• Look again at the answer by Craig. We have $d_{n-2} (f(x_{n-1}) dx_1 \wedge \cdots \wedge dx_{n-2}) = f' (x_{n-1}) dx_{n-1} \wedge dx_1 \cdots \wedge dx_{n-2}$. Unless $f(x_{n-1})$ is constant, you get a non-zero image in $\mathrm{im}(d_{n-2})$, which hence contains at least this infinite dimensional subspace. If you have an explicit form of the $q$ operator, you could probably use the same trick, by looking carefully at its components. – Igor Khavkine Apr 1 '17 at 9:18
• @AsafShachar: Roughly speaking, given a differential equation $Dv = 0$, an integrability condition is any differential consequence $Iv := (C\circ D) v=0$, where the operator $I$ is of the same or lower order than $D$ itself. Example: in your case, let $D=d_0$ and $C=d_1$, note that $(d_1)_j (d_0)_k (f^* v^i) = -f^*(R_{jkl}{}^i v^l)$, where $R_{jkl}{}^i$ is the Riemann tensor on $N$. This integrability condition is of order zero, while $d_0$ was of order $1$. As an exercise, find more integrability conditions, which will involve derivatives of the Riemann tensor. – Igor Khavkine Apr 1 '17 at 9:31

It's clearly infinite dimensional in the flat case.

$\ker\delta_1 =\ker d_{n-1}$

And in a coordinate chart with coordinates $(x_1, \ldots, x_n)$ take various $f\in C^\infty(M)$ with support in the chart, so

$df\wedge dx_1 \wedge\cdots\wedge dx_{n-2} \in \ker d_{n-1}$