Is there any relationship between the convolution and cross-correlation of two signals?
okay, let's associate the symbols $u$ and $v$ with the dimensions of $x$ and $y$ with respect to each.

and let's assume periodicity in both the $x$ and $y$ dimensions for both pictures. $f(x,y)$ and $g(x,y)$.

$$ f(x+m,y) = f(x,y) \qquad \forall x,y $$
$$ f(x,y+m) = f(x,y) \qquad \forall x,y $$
$$ g(x+m,y) = g(x,y) \qquad \forall x,y $$
$$ g(x,y+m) = g(x,y) \qquad \forall x,y $$

then 2-D Cross-Correlation is:

$$ R_{f,g}(u,v) = \sum\limits_{x=0}^{m-1} \sum\limits_{y=0}^{m-1} f(x,y) \cdot g(x+u,y+v) $$

and 2-D circular convolution is:

$$ (f\circledast \hat{g})(u,v) = \sum\limits_{x=0}^{m-1} \sum\limits_{y=0}^{m-1} f(x,y) \cdot \hat{g}(u-x,v-y) $$

if $\hat{g}(x,y) = g(-x,-y)$, then the cross-correlation and the circular convolution are the same.  if $\hat{g}$ is an upside-down copy of $g$, the the correlation and convolution are the same.

or maybe they are upside-down of each other.  i think it's:

if $\hat{g}(x,y) = g(-x,-y)$, then 

$$ (f\circledast \hat{g})(u,v) = R_{f,g}(-u,-v) $$

that looks about right.  someone needs to check this.