Friday, October 31, 2014

A new method of constructing connections on vector bundles


Suppose that $M$ is a smooth manifold of dimension $E\to M$ is a real, smooth vector bundle of rank $\nu$ over $M$.      We define  a pairing on $E$ to be  a section of the bundle $E^*\boxtimes E^*\to M\times M$, where  $E^*\boxtimes E^*$ is the vector bundle $\pi_1^* E^*\otimes \pi_2^*E^*$, $\pi_i(x_1,x_2)=x_i$,  $\forall (x_1,x_2)\in M\times M$, $i=1,2$.

For $x,y\in M$ we can view $B_{x,y}\in E_x^*\times E^*_y$ as a bilinear map

$$B_{x,y}: E_x\times E_y\to \bR.$$

This  induces a  linear map
$$S_{x,y}= S(B)_{x,y}: E_y\to E^*_x.$$
We say that the pairing is nondegenerate if for any $x\in M$ the bilinear map $B(x,x): E_x\times E_x\to \bR$ is nondegenerate. In particular, this induces an isomorphism
$$S_x=S_{x,x}: E_x\to E^*_x.$$

We obtain tunneling operators

$$T(x,y)= S_x^{-1}S_{x,y}: E_y\to E_x.$$


The local frame  $\underline{\be}$ $\newcommand{\ube}{{\underline{\boldsymbol{e}}}}$ defines a bundle isomorphism $\Phi(\underline{\be}):\ur_\eO\to E_\eO$.  In the local frame $\ube$ the tunelling are represented by$\DeclareMathOperator{\Endo}{End}$ $\DeclareMathOperator{\Aut}{Aut}$ a tunneling map

$$T_\ube:\eO\times \eO\to\Endo(\bR^\nu),\;\;T_\ube(x,y)= \Phi_x(\ube)^{-1}T(x,y)\Phi_y(\ube) .$$

Note that $T_\ube(x,x) = \mathbf{1}_{\bR^\nu}$.  For $i=1,\dotsc, m$ define

$$\Gamma_i(\ube):\eO\to \Endo(\bR^\nu),\;\;\Gamma_i(\ube,x)=-\pa_{x^i}T_\ube(x,y)\bigl|_{y=x}.$$

We set

$$\Gamma(\ube,x)=\sum_{i=1}^m \Gamma_i(\ube, x) dx^i=-d_x T(x,y)\bigl|_{y=x}\in \Endo\bigl(\;\bR^\nu\;\bigr)\otimes \Omega^1(\eO),$$
$$\Phi_x(\ubf) =\Phi_x(\ube)\circ g(x),\;\;\forall x\in\eO.$$
Then
$$T_\ubf(x,y)=g(x)^{-1} T_\ube(x,y) g(y),$$
$$\Gamma(\ubf,x) = -d_x\bigl(\; g(x)^{-1}\;\bigr)\bigl|_{y=x} T_\ube(x,x)g(yx+g^{-1}(x) \Gamma(\ube,x) g(x)=g(x)^{-1}dg(x)+g^{-1}(x) \Gamma(\ube,x) g(x).$$

This proves that the correspondence $\ube\mapsto \Gamma(\ube)$ defines a connection on $E$. We will denote it by $\nabla^B$ and we will refer to it as the connection  associated to the nondegenerate pairing $B$.

Let us compute its  curvature $R^B$.  Using the local frame $\ube$ we can write
$$R^B= \sum_{1\leq i<j\leq m} R_{ij}(\ube, x)dx^i\wedge dx^j\in \Endo(\bR^\nu)\otimes\Omega^2(\eO),$$
where
$$R_{ij}(\ube,y)=\pa_{x^i}\Gamma_j(\ube,x)-\pa_{x^j}\Gamma_i(\ube,x)+[\Gamma_i(\ube,x),\Gamma_j(\ube,x)].$$

Using the local frame $\ube$ we represent $S_{x,y}$ as a $\nu\times \nu$-matrix

$$S_\ube(x,y)=\bigl(\; s_{\alpha\beta}(x,y)\,\bigr)_{1\leq\alpha,\beta\leq \nu},\;\;s_{\alpha\beta}(x,y)=B_{x,y}\bigl(\,\be_\alpha(x),\be_\beta(y)\;\bigr).$$
Then $T_\ube(x,y)=S_\ube(x,x)^{-1} S_\ube(x,y)$, and
$$\Gamma_i(\ube,x)= -\pa_{x^i} S_\ube(x,x)^{-1}\bigl|_{y=x} S_\ube(x,x)-S_\ube(x,x)^{-1} \pa_{x^i}S_\ube(x,y)\bigl|_{y=x}$$

= S_\ube(x,x)^{-1}\pa_{x^i} S_\ube(x,x)\bigl|_{x=y}-S_\ube(x,x)^{-1}  \pa_{x^i}S_\ube(x,y)\bigl|_{y=x} =S_\ube(x,x)^{-1}  \pa_{y^i}S_\ube(x,y)\bigl|_{y=x}.\label{gamma}

We have
$$\pa_{x^i}\Gamma_j(\ube,x)=\pa_{x^i}\Bigl(\; S_\ube(x,x)^{-1} \pa_{y^j}S_\ube(x,y)\bigl|_{y=x}\;\Bigr)$$

$$= -S_\ube(x,x)^{-1}\Bigl(\pa_{x^i}S_\ube(x,x)\;\Bigr) S_\ube(x,x)^{-1} \pa_{y^j}S_\ube(x,y)\bigl|_{x=y}+ S_\ube(x,x)^{-1}\pa_{x^i}\Bigl( \; \pa_{y^j}S_\ube(x,y)\bigl|_{x=y}\;\Bigr)$$

=- S_\ube(y,y)^{-1}\Bigl(\pa_{x^i}S_\ube(x,y)+\pa_{y^i}S_\ube(x,y)\;\Bigr)_{x=y}S_\ube(y,y)^{-1}   \pa_{y^j}S_\ube(x,y)\bigl|_{x=y} +S_\ube(y,y)^{-1} \pa^2_{x^iy^j}S_\ube(x,y)\bigr|_{x=y}. \label{1}

We can simplify the computations a bit if we choose the frame  $\ube$ judiciously.  Fix a distinguished point in $\eO$ and assume it is the orgin in the coordinates $(x^i)$. Note that if $x$ is sufficiently close to $0$, then $T(x,0)$ is an isomorphism $E_0\to E_x$. We set

$$\bsf_\alpha(x): = T(x,0)\be_\alpha(0).$$

More explicitly

$$\bsf_\alpha(x)=\sum_{\gamma,\lambda} s^{\gamma\lambda}(x)s_{\lambda \alpha}(x,0)\be_\gamma(x),$$

where $(s^{\gamma\lambda}(x))$ is the inverse of the matrix $(s_{\alpha\beta}(x) )$. $\newcommand{\one}{\mathbf{1}}$

In this frame we have  $T_\ubf(x,0)=\one$ and we deduce that

\Gamma(\ubf, 0)=0.
\label{2}

On the other hand,
$\Gamma_i(\ubf,0)=S_\ubf(0,0)^{-1}\pa_{y^i}S_\ubf(0,y)\bigr|_{y=0}.$
We deduce that for this special frame we have
$\pa_{y^i}S_\ubf(0,y)\bigr|_{y=0}=0.$
Using this in (\ref{1}) we deduce

\pa_{x^i}\Gamma_j(\ubf,0)=S_\ubf(0,0)^{-1} \pa^2_{x^iy^j}S_\ubf(0,0),
\label{3}

and thus

R_{ij}(\ubf,0)=S_\ubf(0,0)^{-1}\Bigl(\pa^2_{x^iy^j}S_\ubf(0,0)-\pa^2_{x^jy^i}S_\ubf(0,0)\Bigr).
\label{curv}

Remark.    We say that the pairing $B$ is symmetric  if  for any $x,y\in M$ and any $u\in Y_x$, $v\in E_y$ we have
$B_{x,y}(u,v)=B_{y,x}(v,u).$
Observe that the symmetry  condition is equivalent to requiring that  the tunneling $T_{x,y}: E_y\to E_x^*$ is  self-adjoint, i.e., the adjoint $T_{x,y}^*: (E_x^*)^*\to E_y^*$ coincides with $T_{y,x}$.

In this case  it is not easy to prove that the bilinear form $\hat{B}\in C^\infty(E^*\otimes E^*)$, $\hat{B}_x=B_{x,x}$, is  covariant constant

\nabla^B \hat{B}=0.
\label{const}

Example. Let us investigate a special case when $E=TM$ and the pairing $B$ has the special form

$$B^f_{x,y}(X,Y)= XYf(x,y),\;\;X\in T_xM,\;\;Y\in T_yM,$$

where $f\in C^\infty(M\times M)$ is a given function.  Note that $f$ has to be a rather special function so that $B^f_{x,x}$ is nondegenerate. (This is the context of the MathOverflow question that prompted this post.) We thus obtain a connection $\nabla^f$ on $TM$.

Once we choose local coordinates $(x^i)$ on $\eO$ we have a natural frame
$\be_\alpha=\pa_\alpha:=\pa_{x^\alpha}.$
With these choices $S_{x,y}$ is represented by the matrix
$s_{\alpha\beta}(x,y)=\pa{x^\alpha}\pa_{y^\beta}f(x,y).$
We denote by $s^{\alpha\beta}(x)$ the inverse of $(s_{\alpha\beta}(x,x))$. Then $\Gamma_i(x)$ is an $m\times m$ matrix
$\Gamma_i(x)=\bigl(\;\Gamma^\alpha_{\beta i}(x)\;\bigr)_{1\leq\alpha,\beta\leq m}.$
Using (\ref{gamma}) we deduce that
$\Gamma^\alpha_{\beta i}(x)= \sum_{\gamma} s^{\alpha\gamma}(x) \pa^3_{x^\gamma y^\beta y^i}f(x,x).$
Note that this implies that
$\Gamma^\alpha_{\beta i}=\Gamma^\alpha_{i\beta},$
which shows that in this case   we get a torsion free connection. The pairing $B^f$ is  symmetric when $f$ is symmetric, i.e., $f(x,y)=f(y,x)$, $\forall x,y\in M$.   Thus, when $f$ is symmetric   the nondegenerate symmetric tensor $\hat{B}^f\in C^\infty(T^*M^{\otimes 2})$ is covariant constant with respect to this connection.

Here is a special case of this construction. Suppose that $M$ is a submanifold of $\bR^N$.  The function
$F:\bR^N\times \bR^N\to \bR,\;\;F(x,y)= x\cdot y,$
restricts to a function $f$ on $M\times M$ satisfying the above nondegneracy condition. In this case, for $x\in M$ the bilinear form $B^f_{x,x}$ on $T_xM$ is the inner product induced by the canonical inner product on $\bR^N$.  The  forms $B^f_{x,x}$ define a Riemann metric on $M$ and, as explained  in  my paper,  the connection $\nabla^f$ is compatible with this metric. Since  it is torsion free, it must be the Levi-Civita connection.