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Let $M$ be a compact, finite-dimensional manifold and $\pi : \mathrm{B} \rightarrow M$ a vector bundle over $M$ whose typical fiber is $\mathbb{R}^n$. Denote by $\mathcal{C}^{\infty}(M, \mathrm{B})$ the vector space of all smooth sections of $\mathrm{B}$.

If we choose Riemannian metrics on (the fibers of) $\mathrm{B}$ and on $M$, and then introduce a metric connection $D$ on $B$, we may define a family of seminorms on $\mathcal{C}^{\infty}(M, \mathrm{B})$ by $$\| s \|_n = \sum_{i = 0}^n~\sup_{x \in M} |D^js(x)|,$$ where $|D^0s(x)|$ is just $|s(x)|$, and for $j \geq 1$, $$|D^js(x)| = \sup |(D_{v_1} \circ \dots \circ D_{v_j}s)(x)|,$$ the supremum being taken over all $(v_1, \ldots, v_j) \in (T_xM)^j$ with $|v_k| = 1$, for $k = 1, \ldots, j$. It's not hard to prove that if $\{s_n\}_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathcal{C}^{\infty}(M, \mathrm{B})$ with respect to this family of seminorms, then it converges to a continuous section $s : M \rightarrow \mathrm{B}$. I'd like to prove that $s$ is actually smooth.

QUESTION: What would be the best ($\sim$ least messy, shortest) way to prove this?


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Isn't that obvious if you locally pick an orthonormal frame $E_i$ for $B$ and a geodesic chart, giving an orthonormal basis $\partial_j$ for $TM$? – Blah Mar 25 '12 at 21:33

Are you sure this is true?

Consider the following: take $M = S^1$ to be the unit circle and take $B \to M$ to be trivial line bundle with fiber $\mathbb R^1$. The space of smooth global sections of $B$ is then just the space of smooth functions on $S^1$. Equip $S^1$ with the usual metric on the circle and equip $B$ with the Euclidean metric. Take the connections involved to be the trivial ones.

Now let $f : S^1 \to [-1,1]$ be the function induced by $x \mapsto \sin(2\pi x)$. This function is smooth, and thus a smooth section of $B$. Note that the absolute value function $|\cdot| : [-1,1] \to \mathbb R$ is continuous. By the Weierstrass approximation theorem there exists a sequence of polynomials $p_m$ that converges uniformely to $|\cdot|$ on $[-1,1]$. The composition $s_m = p_m \circ f$ is thus a sequence of smooth sections of $B$, that converges to $s(x) = |\sin(2\pi x)|$, which is not a smooth section of $B$ (its derivative is discontinuous at the point corresponding to the zero $\pi$ - just plot the graph of the function to see it).

I haven't verified the details, so it is possible that the sequence $(s_m)$ does not converge in the seminorm required (basically due to the discontinuity of the first derivative of the limit). If so, then replacing the absolute value by a function with continuous first, but not second, derivatives should give a counterexample.

In general I believe one obtains the space of $L^2$ sections of $B$ by taking the completion with respect to the seminorm you defined, and not the space of smooth sections.

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I think that your sequence really does not converge in the $\|~\|_1$ seminorm, as you said. Moreover, replacing the absolute value by a function with $k$ continuous derivatives but such that its $(k+1)$-derivative is not continuous won't work because convergence in the $\|~\|_{k+1}$ seminorm will fail. Note that this topology is the one induced by all seminorms, not just a fixed $n$. What do you think? – student Feb 24 '12 at 17:45
Aahhh, merde, you're right. I misread and thought you only had one seminorm in play and not a family of them. In that case I'd try to reduce to a relatively compact open set in a trivializing neighborhood of a point and see what that gives. I don't remember my functional analysis all that well, but is it clear in the local case that the limit is smooth and not just $L^2$ or a distribution? – Gunnar Þór Magnússon Feb 24 '12 at 18:20
Well, yes.. in the local case I believe it's more or less clear, although I haven't checked the details. I was hoping there would be a shorter proof, but perhaps there isn't. – student Feb 25 '12 at 13:10

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