I am currently studying this paper (page 53), in which the suggest convolution to be done in a special manner.
This is the formula:
\begin{equation} \label{eq:equation} q_{j,m} = \sigma \left(\sum_i \sum_{n=1}^{F} o_{i,n+m-1} \cdot w_{i,j,n} + w_{0,j} \right) \end{equation}
Here is their explanation:
As shown in Fig. 4.2, all input feature maps (assume I in total), $O_i (i = 1, · · · , I)$ are mapped into a number of feature maps (assume $J$ in total), $Q_j (j = 1, · · · , J)$ in the convolution layers based on a number of local filters ($I × J$ in total), $w_{ij}$ $(i = 1, · · · , I; j = 1, · · · , J)$. The mapping can be represented as the well-known convolution operation in signal processing. Assuming input feature maps are all one dimensional, each unit of one feature map in the convolution layer can be computed as equation \ref{eq:equation} (equation above).
where $o_{i,m}$ is the $m$-th unit of the $i$-th input feature map $O_i$, $q_{j,m}$ is the $m$- th unit of the $j$-th feature map $Q_j$ of the convolution layer, $w_{i,j,n}$ is the $n$th element of the weight vector, $w_{i,j}$, connecting the $i$th feature map of the input to the $j$th feature map of the convolution layer, and $F$ is called the filter size which is the number of input bands that each unit of the convolution layer receives.
So far so good:
What i basically understood from this is what I've tried to illustrate in this image.
It seem to me what they are doing is actually processing all data points up to F, and across all feature maps. Basically moving in both x-y direction, and compute on point from that.
Isn't that basically 2d- convolution on a 2d image of size $(I x F)$ with a filter equal to the image size?. The weight doesn't seem to differ at all have any importance here..?
As a partial answer the definition of convolution is wrong. The real definition is in the continous case: $$(f*g)(t) = \int_{-\infty}^{\infty} f(\tau)g(t-\tau)d\tau$$ and in the discrete case:
$$(f*g)[n] = \sum_{k=-\infty}^{\infty} f[k]g[n-k]$$
You can clearly see that $\tau$ and $k$ both have opposite sign in the function arguments, which correspond to indexes in your notation.