# Harris corner detection shape of $E(u,v)$

I am taking a computer vision class and I have just learnt about the Harris corner detection concept. A corner is detected when a small shift in a window function defined around the corner results in a large $$E(u,v)$$ term, which is the sum of squared difference of pixels between the previous window and the next window. There is a post on this forum that has a very nice conceptual explanation about harris corner detectors.

However, given that $$E(u,v)$$ should be large for small changes in $$(u,v)$$, why is $$E(u,v)$$ function usually depicted as a function that has a negative maxima ? Shouldn't $$E(u,v)$$ value be high when $$(u,v)$$ is close to $$(0,0)$$  $$E(u,v)$$ is zero for zero shift ($$u=v=0$$), where the two patches being compared are equal. As you increase the distance, the difference tends to increase. Depending on the local shape of the image, the $$E(u,v)$$ surface will have a different shape: at a corner or a small point, it will steeply increase from 0 in all directions. Along an edge, it will remain close to 0 along one direction, and increase steeply in the perpendicular direction. At flat image regions, the function will remain closer to zero in all directions.