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For a Hermitian vector bundle E over a Riemannian $(X,g)$ and a unitary connection A we may define the sup norm of the curvature by: $$\|R_A\|=\sup \{\| tracefree (R_A (v))\|_{op} \mid v \in \Lambda ^2 TX, \,\|v\|_g=1 \},$$ where $\| \cdot \|_{op}$ is the operator norm, and tracefree denotes the traceless part of the operator. Define $$K (E)= \inf_A \|R_A\|.$$

Question: For $X = S^{2k}$, $k>1$, $E$ having non vanishing Chern number, does $K(E^n)$ eventually grow linearly with $n$ ($E^n$ is the $n$ fold direct sum here)? Note that Chern-Weil theory does imply growth but not linear growth.

It (seems) to be possible to show that this is implied by existence of universal norm minimizing connection on the universal $\mathbb{C}^r$ bundle $\mathcal {E}$ over $BU(r)$ (this is considered as a Riemannian manifold in a diffeological sense, or simply as a direct limit of Riemannian manifolds). Specifically we want $\cal{A}$ a connection on $\mathcal {E}$, with constant curvature, s.t. a connection $A$ on $E$ minimizing the norm, is gauge equivalent to $(f^* \mathcal {E}, f^* \mathcal A)$, for some $f: X \to BU(r)$. This could be a harder question than original, but is more conceptual.

Edit: added tracefree, changed k to 2k.

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  • $\begingroup$ There is a paper by Narasimhan and Ramanan that answers your second question - every connection can be pulled back from the standard connection on the tautological bundle over the Grassmann manifold $G_{r,N}(\mathbb C)$, provided $N$ is large enough ($N$ is determined by $r$ and $\dim X$ alone, if I remember correctly). $\endgroup$
    – Sebastian Goette
    Commented Dec 7, 2015 at 18:53
  • $\begingroup$ Btw, have you read Gromov's article? Somewhere in section 4-6 he does something looking very similar to your question. $\endgroup$
    – Sebastian Goette
    Commented Dec 8, 2015 at 14:00
  • $\begingroup$ Sebastian the original version of the question was also ok, since the operator norm if the direct sum connection is exactly double. Adding tracefree does not substantially change the question, but I keep this change since ultimately this is what I want. $\endgroup$
    – Yasha
    Commented Dec 10, 2015 at 14:30
  • $\begingroup$ I think this is just a difference of what we mean by operator norm, for me this is: $\mathfrac{||Av}||}{||v||}$, for a direct sum operator the operator norm adds. $\endgroup$
    – Yasha
    Commented Dec 11, 2015 at 15:00
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    $\begingroup$ Actually tracefree does not really help since at least in the formulation above one could restrict wlog to SU(n) connections. So at least for the formulation above the answer looks to be: no there is no growth. $\endgroup$
    – Yasha
    Commented Dec 11, 2015 at 19:00

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