Return to Question

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-Weyl 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.

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.

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-Weyl 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.

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-Weyl 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.

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: $$\|R_A\|=\sup \{\| R_A (v)\|_{op} \mid v \in \Lambda ^2 TX, \,\|v\|_g=1 \},$$ where $\| \cdot \|_{op}$ is the operator norm. 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-Weyl 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.

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-Weyl 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.