Preservation of the curvature tensor implies preservation of the connection?

Question

For every connection on a smooth manifold there is a corresponding curvature tensor. Any diffeomorphism $\phi:(M,\nabla^M)\rightarrow(N,\nabla^N)$ which preserves the connection (in the sense of $\nabla^N_X{Y}=\phi_* \left( \nabla^{M}_{\phi^{-1}_*(X)} {\phi^{-1}_*(Y)} \right)$ ) preserves the curvature tensor.

My question is about the "reverse direction":

Does every diffeomorphism which preserves the curvature preserves the original connection behind it? I suspect the answer is negative but I could not come up with an example.

Answer 1

There are many different connections on the same manifold which share the same curvature tensor.

For example, take a connection $\nabla$. If $A$ is any global positive function $M\to R$, consider the connection $\tilde \nabla$ given by:

$$\tilde \nabla_X Y := \nabla_X Y + X(A)Y \;.$$

It is linear on both entries, and: $$\tilde \nabla_X (fY) = \nabla_X (fY) + X(A)\,fY = X(f) + f\tilde \nabla_X (Y), $$

so it is indeed a connection. Its curvature is given by: $$ \tilde R(X,Y)\,Z = [\tilde\nabla_X,\tilde\nabla_Y]\,Z - \tilde\nabla_{[X,Y]}Z. $$

If you insert the expression for $\tilde \nabla_X Y$, the terms in $A$ cancel out, giving you exactly the same curvature as $\nabla$.

Answer 2

Another approach would be to focus upon connections whose curvature tensor is trivial. For example, take $M=\mathbb{R}^n$ with $\nabla$ the Levi-Civita connection w.r.t Euclidean metric.

As I prove below, $Iso(\mathbb{R},\nabla)=E(n)$, the group of affine transformations. But $R(\nabla)=0$ so any diffeomorphism of $\mathbb{R}$ preserves it. Now, just choose some non-affine diffeomorphism of $\mathbb{R}$, and you are done.

(For a concrete example: $f(x)=2x+sin(x)$. $f$ is smooth, $f'(x) \ge 1$ so $f$ is strictly increasing, hence it is a a smooth bijection. Now by the inverse function theorem, it is actually a diffeomorphism).

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Theorem: $Iso(\mathbb{R},\nabla)=E(n)$ the group off affine transformations. (where $\nabla$ is the Levi-Civita connection of the Euclidean metric).

Proof:

Take $X=\frac{\partial}{\partial t},Y=g\frac{\partial}{\partial t}$ where $f,g \in C^\infty(\mathbb{R})$ and note that in general: $\phi_*(f \cdot X)=(f \circ \phi^{-1}) \cdot \phi_*(X)$

$ \nabla_{\phi_*X}({\phi_*Y})=\phi_* (\nabla_X Y) \iff \nabla_{\phi_*\frac{\partial}{\partial t}}({\phi_*[g\frac{\partial}{\partial t}}]) = \phi_* (\nabla_\frac{\partial}{\partial t} g\frac{\partial}{\partial t}) = \phi_* (g'(t)\frac{\partial}{\partial t})$

Now using the fact: $(\phi_*\frac{\partial}{\partial t}) = (\phi' \circ \phi^{-1}) \cdot \frac{\partial}{\partial t}$, we get:

$\iff \nabla_{(\phi' \circ \phi^{-1}) \cdot \frac{\partial}{\partial t}}({\phi_*[g\frac{\partial}{\partial t}}]) = \phi_* (g'(t)\frac{\partial}{\partial t}) \iff$

$ (\phi' \circ \phi^{-1}) \cdot \nabla_\frac{\partial}{\partial t} ({\phi_*[g\frac{\partial}{\partial t}}]) = (g' \circ \phi^{-1}) \cdot \phi_*\frac{\partial}{\partial t} \iff$

$ (\phi' \circ \phi^{-1}) \cdot \nabla_\frac{\partial}{\partial t} [(g \circ \phi^{-1})\cdot (\phi_*\frac{\partial}{\partial t})] = (g' \circ \phi^{-1}) \cdot \phi_*\frac{\partial}{\partial t} \iff$

$ (\phi' \circ \phi^{-1}) \cdot \nabla_\frac{\partial}{\partial t} [(g \circ \phi^{-1}) \cdot (\phi' \circ \phi^{-1}) \cdot \frac{\partial}{\partial t}] = (g' \circ \phi^{-1}) \cdot (\phi' \circ \phi^{-1}) \cdot \frac{\partial}{\partial t} \iff$

$\nabla_\frac{\partial}{\partial t} [\left( (g \cdot \phi' ) \circ \phi^{-1} \right) \frac{\partial}{\partial t} ] = (g' \circ \phi^{-1}) \cdot \frac{\partial}{\partial t} \iff$

$\left( (g \cdot \phi' ) \circ \phi^{-1} \right)' = g' \circ \phi^{-1} \iff [(g \cdot \phi')' \circ \phi^{-1}] \cdot (\phi^{-1})' = g' \circ \phi^{-1} \iff $

$ g' \circ \phi^{-1} = \left( (g' \cdot \phi' + g \cdot \phi'') \circ \phi^{-1} \right) \cdot (\phi^{-1})' = (g' \circ \phi^{-1}) \cdot (\phi \circ \phi^{-1})' + [(g \cdot \phi'') \circ \phi^{-1} ] \cdot (\phi^{-1})' = g' \circ \phi^{-1} + [(g \cdot \phi'') \circ \phi^{-1} ] \cdot (\phi^{-1})' \iff$

$[(g \cdot \phi'') \circ \phi^{-1} ] \cdot (\phi^{-1})' = 0 \iff (g \cdot \phi'') \circ \phi^{-1} = 0 \iff g \phi'' = 0$. Since this must hold for every function $g$, it follows that $\phi''=0$.

Therfore, $\phi$ is affine as required. (Actually we only showed that any preserving map must be affine. But the other direction is easy).

Answer 3

(This solution is based on the idea of geodude).

Let $M$ be a manifold with a connection $\nabla$, and let $A \in C^\infty (M)$.
Define another connection: $\tilde \nabla_X Y = \nabla_X Y + X(A)Y.$

We need two technical lemmas (proofs are provided below).

Lemma $\text{1}$: $R^{\nabla} = R^{\tilde \nabla}$

Lemma $\text{2}$: Let $\phi \in \text{Iso}(M,\nabla)$. Then $\phi \in \text{Iso}(M,\tilde \nabla) \iff dA_q \circ d\phi_{\phi^{-1}(q)} = dA_{\phi^{-1}(q)}$.

Here is the idea:
Since $\phi \in \text{Iso}(M,\nabla) \Rightarrow \phi \in \text{Iso}(M,R^{\nabla}) \stackrel{\mathrm{Lemma 1}}{=} \text{Iso}(M,R^{\tilde \nabla})$, it is enough to find an isomorphism of $\nabla$, which is not an isomorphism of $\tilde \nabla$. By Lemma 2, we have an explicit condition which is easy to violate, thus finding a suitable isormophism as required.

A concrete example:
$M= \mathbb{R},\nabla $ the Levi-Civita connection w.r.t the standard metric on $\mathbb{R}$, $\phi(q) = -q = \phi^{-1}(q)$ is an isometry, hence it preserves $\nabla$. For a given function $A: \mathbb{R} \rightarrow \mathbb{R}$, $\phi$ preserves $\tilde \nabla^A$ iff $-dA_q = dA_{-q} \iff A'(-q)=-A'(q)$, but there are many functions $A$ which do not satisfy this.

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Proof of Lemma $\text{2}$:

$\tilde \nabla_{\phi_*X}({\phi_*Y})=\phi_* (\tilde \nabla_X Y) \iff \nabla_{\phi_*X}({\phi_*Y}) + [(\phi_*X)(A)] \cdot\phi_*Y = \phi_* (\nabla_X Y) + \phi_* (X(A)Y) \stackrel{\mathrm{\phi preserves \nabla}}{\iff} [(\phi_*X)(A)] \cdot\phi_*Y = \phi_* (X(A)Y ) \iff [(\phi_*X)(A)] \cdot\phi_*Y = (X(A)\circ \phi ^{-1}) \cdot\phi_*Y$

Since $Y$ and hence $\phi_*Y$ can be arbitrary vector fields on $M$, and in particular for every $q \in M, \left(\phi_*Y \right)(q)$ can be a non-zero vector in $T_qM$ , this forces $(\phi_*X)(A) = X(A)\circ \phi ^{-1}$.

So $\phi \in \text{Iso}(M,\tilde \nabla) \iff \forall q \in M,X \in \Gamma(TM) , (\phi_*X)(A)(q) = X(A)\circ \phi ^{-1}(q) \iff dA_q((\phi_*X)(q)) = dA_{\phi^{-1}(q)}\left( X(\phi^{-1}(q)) \right ) \iff dA_q\left(d\phi_{\phi^{-1}(q)}(X\left(\phi^{-1}(q)\right)) \right) = dA_{\phi^{-1}(q)}\left( X(\phi^{-1}(q)) \right )$.
Since $X$ is arbitrary, this forces: $dA_q \circ d\phi_{\phi^{-1}(q)} = dA_{\phi^{-1}(q)}$.

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Proof of Lemma $\text{1}$:

$\tilde R(X,Y)\,Z = [\tilde\nabla_X,\tilde\nabla_Y]\,Z - \tilde\nabla_{[X,Y]}Z = \tilde\nabla_X (\tilde\nabla_Y Z)-\tilde\nabla_Y (\tilde\nabla_X Z)-\tilde\nabla_{[X,Y]}Z = \tilde\nabla_X (\nabla_Y Z+Y(A)Z)-\tilde\nabla_Y (\nabla_X Z+X(A)Z)-\big[\nabla_{[X,Y]}Z+\big([X,Y]A\big)Z\big] = [\nabla_X (\nabla_Y Z+Y(A)Z) + X(A)(\nabla_Y Z+Y(A)Z)]-[\nabla_Y (\nabla_X Z+X(A)Z)+Y(A)(\nabla_X Z+X(A)Z)]-\big[\nabla_{[X,Y]}Z+\big([X,Y]A\big)Z\big] = $

$ R(X,Y)Z + \nabla_X(Y(A)Z) + X(A)(\nabla_Y Z+Y(A)Z) - \nabla_Y (X(A)Z)-Y(A)(\nabla_X Z+X(A)Z)-\big([X,Y]A\big)Z$

So,

$\tilde R(X,Y)\,Z - R(X,Y)Z = X(Y(A))Z+Y(A)\nabla_X Z + X(A)(\nabla_Y Z+Y(A)Z) - Y(X(A))Z - X(A) \nabla_Y (Z)-Y(A)(\nabla_X Z+X(A)Z)-\big([X,Y]A\big)Z = 0.$