# Does existence of a parallel section on $E$ imply local existence of a parallel section on $\bigwedge^k E$?

Let $$E$$ be a smooth vector bundle equipped with an affine connection $$\nabla$$.

Suppose that $$(E,\nabla)$$ admits a non-zero parallel section. I think that $$(\bigwedge^k E,\bigwedge^k \nabla)$$ does not need to admit a non-zero parallel section even locally. How to construct such an example?

This seems especially interesting when $$k < \text{rank}(E)$$.

Moreover generally , are there any non-trivial relations between the dimension of the space of parallel sections of $$(E,\nabla)$$, and that of $$(\bigwedge^k E, \bigwedge^k\nabla)$$ (locally)?

If there are $$k$$ independent parallel sections of $$E$$, then $$\bigwedge^k E$$ has at least one parallel section; $$\sigma_1,\dots,\sigma_k$$ are parallel $$\Rightarrow \sigma_1 \wedge \dots \wedge \sigma_k$$ is parallel.

What happens if $$E$$ has $$r parallel sections? Does $$\bigwedge^k E$$ still admit (locally) a non-zero parallel section?

Edit: We can probably use the relation between the curvatures: Let $$X,Y \in \Gamma(TM)$$. Then $$R^{ \bigwedge^k\nabla}(X,Y)=d\psi_{\operatorname{Id}} (R^{\nabla}(X,Y))$$, where $$\psi:\text{End}(E) \to \text{End}(\bigwedge^k E)$$ is the exterior power map, $$\psi(A)=\bigwedge^k A$$.

Since the dimension of the space of local parallel sections around $$p \in M$$ equals $$\ker R(\cdot,\cdot)$$, it suffices to construct an example where $$R^{\nabla}(\cdot,\cdot)$$ is singular, but $$R^{ \bigwedge^k\nabla}(\cdot,\cdot)$$ is invertible. This is certainly possible on the algebraic level, see e.g. example 1, in this question, with $$k=2, \text{rank}E=3$$.

• So, certainly when $k=\text{rank}(E)$ you can't have a nowhere-vanishing section of $\Lambda^k E$ unless $E$ is an orientable bundle. I suspect you can create examples in general by taking $E=F\oplus L$ with $L$ a trivial line bundle (with an obvious split connection). – Ted Shifrin Feb 10 at 22:19
• Thank you. I am actually more interested in the local problem, but I forgot to mention this in the question. My apologies. (I have now edited the question accordingly). – Asaf Shachar Feb 11 at 7:57

Fix a manifold $$M$$ of dimension $$m \geq 2$$ and let $$E$$ be a trivial rank two bundle over $$M$$. Since your question is local, we might as well assume that $$M$$ is just an open subset of $$\mathbb{R}^m$$ but this won't simplify anything. Let $$(e_1,e_2)$$ be a global frame for $$E$$ over $$M$$ and choose some non-closed one form $$\alpha \in \Omega^1(M)$$. Define a connection $$\nabla = \nabla^{\alpha}$$ on $$E$$ by the formula

$$\nabla_X(f^1 e_1 + f^2 e_2) = (Xf^1)e_1 + ((Xf^2) + \alpha(X)f^2)e_2. \tag{1}$$

You can readily check that this indeed defines a connection and it satisfies

$$\nabla_X(e_1) \equiv 0, \nabla_X(e_2) = \alpha(X) e_2,\\ R_{\nabla}(X,Y)(f^1 e_1 + f^2 e_2) = f^2 \cdot d\alpha(X,Y) \cdot e_2.$$

In particular, $$R_{\nabla} \neq 0$$ so $$E$$ doesn't have two independent parallel sections with respect to $$\nabla$$, only one (up to a constant multiple): $$e_1$$. Let's check whether $$\Lambda^2(E)$$ has a non-trivial parallel section. A general section of $$\Lambda^2(E)$$ has the form $$g(e_1 \wedge e_2)$$ where $$g$$ is a smooth function. Then

$$\nabla_X(g(e_1 \wedge e_2)) = (Xg)(e_1 \wedge e_2) + g(\nabla_X(e_1 \wedge e_2)) = (Xg)(e_1 \wedge e_2) + g(\nabla_X e_1 \wedge e_2 + e_1 \wedge \nabla_X e_2) = (Xg + g\cdot \alpha(X))e_1 \wedge e_2.$$

Hence, $$\Lambda^2(E)$$ has a non-trivial parallel section if and only if we can find a non-zero $$g$$ which satisfies the equation

$$dg + g \alpha = 0. \tag{2}$$

This is a partial differential equation for $$g$$. Wedging the equation with $$\alpha$$ and taking into account that $$\alpha$$ is a one-form, we get $$dg \wedge \alpha + g \alpha \wedge \alpha = dg \wedge \alpha = 0.$$ Taking the exterior derivative of the equation, we get

$$0 = d^2g + dg \wedge \alpha + g d\alpha = g d\alpha.$$

Since we are looking for non-zero $$g$$, we see that a necessary condition for the equation to have a solution is $$d\alpha = 0$$ which doesn't hold since we assumed $$\alpha$$ is not closed.

In fact, for the family of connections $$\nabla^{\alpha}$$ we see that $$\nabla^{\alpha}$$ has two linearly independent parallel sections if and only if the induced connection on $$\Lambda^2(E)$$ has one linearly independent parallel section. The curvature of the induced connection on $$\Lambda^2(E)$$ can be identified simply with $$d\alpha$$. Indeed, $$R^{ \bigwedge^2\nabla}(X,Y)(e_1 \wedge e_2)=R^{ \nabla}(X,Y)(e_1) \wedge e_2+e_1 \wedge R^{ \nabla}(X,Y)(e_2)=d\alpha(X,Y)e_1 \wedge e_2.$$

Since $$\Lambda^2(E)$$ is a line bundle, it admits a non-zero parallel section if and only if its curvature tensor $$R^{ \bigwedge^2\nabla}$$ vanished identically, i.e. if and only if $$\alpha$$ is closed.

Also, note that formula $$(1)$$ implies that $$Xg + g\cdot \alpha(X)=0$$ if and only if $$\nabla_X(g e_2)=0$$. This gives an alternative proof for the fact that equation $$(2)$$ admits a non-zero solution if and only if $$\nabla$$ is flat.

No, this is far from true, even in the special case of connections $$\nabla$$ on the tangent bundle $$E = TM$$.

To construct an explicit example, take flowbox coordinates adapted to the $$\nabla$$-parallel vector field $$X$$, i.e., so that $$X = \partial_x$$, which imposes exactly that the Christoffel symbols satisfy $$\Gamma_{xj}^\ell = 0$$ for all $$j, \ell$$. In the case $$\dim M = 2$$, substituting these identities in the formula from this previous answer of mine to a related question of yours shows that the curvature of the connection $$\nabla^2$$ induced on $$\Lambda^2 TM$$ is $$R^2 = (-\partial_y \Gamma_{xy}^y + \partial_x \Gamma_{yy}^y) dx \wedge dy ,$$ but that answer also shows that $$\Lambda^2 TM$$ admits a parallel section iff $$R^2 = 0$$, and we can choose $$\Gamma_{xy}^y, \Gamma_{yy}^y$$ for which $$R^2 \neq 0$$. It follows quickly from that this an analogous conclusion applies to $$\bigwedge^n T(M \times \Bbb R^{n - 2})$$, giving counterexamples for arbitrary dimension $$k = \operatorname{rank} E \geq 2$$. With some CAS help it's easy to produce counterexamples for $$E = TM$$ and $$1 < k < \dim M$$, too. (Of course, these cases are more computationally involved, as the curvatures there are sections of $$\Lambda^2 T^*M \otimes \operatorname{End} \Lambda^k E$$, and for $$0 < k < \operatorname{rank} E$$ we have $$\operatorname{rank} \operatorname{End} \Lambda^k E > 1$$.)

There are special cases, however, in which the claim is true. For example (working locally) on a Riemannian manifold $$(M, g)$$, an orientation determines a Hodge star operator $$\ast$$ on multivectors, and since any metric connection $$\nabla$$ preserves $$g$$ and the orientation, it preserves $$\ast$$, too. Thus, if $$X$$ is a nonzero parallel section of $$TM$$, $$\ast X$$ is a nonzero section of $$\bigwedge^{k - 1} T^*M$$ parallel with respect to the connection induced by $$\nabla$$ on that bundle.