Property of Jacobian Determinant

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We can denote the jacobian of a vector map ##\pmb{g}(\pmb{x})## by ##\nabla \pmb{g}##, and we can denote its determinant by ##D \pmb{g}##. We were asked to prove that

##\sum_j \frac{\partial ~ {cof}(D \pmb{g})_{ij}}{\partial x_j} = 0##

generally holds so long as the ##g_i## are suitably differentiable not too long ago. Looking around, this property is mentioned in passing in papers here and there. Does this result go by a headword that I can look up. Like a theorem name of some sort? It is rumored to be an essential property for things like degree theory and the brouwer fixed point theorem.
 
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mathwonk
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i can't really understand your notation, I assume cof stands for cofactor, but since to me a determinant is a number, hence does not have a cofactor, rather a matrix does. I have never seen this inequality in my study of either degree theory or brouwer's fix point theorem, and do not really know what it is saying, but just off the top of my head it looks as if it might follow from the equality of mixed 2nd partials. you might try working it out for n = 2 or 3.

well here is a discussion that gives a reason for it: see p. 2: he calls the general formula he is deriving Jacobi's formula, but this is just a step in the derivation.

https://www.impan.pl/~pmh/teach/algebra/additional/jacobi.pdf

also see page 6 in these notes:

https://www.gotohaggstrom.com/Jacobi’s formula for the derivative of a determinant.pdf
 
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You...are absolutely correct, my friend. I meant
## \sum_j \frac{ \partial ~ cof( \nabla \pmb{g} ) }{ \partial x_j } = 0 ~.##​
But that aside, the above statement is true. I had already proved it, so I was inquiring its significance. For example, I've seen it in the second page (pg 17) of the following link, The Jacobian Determinant Revisited. Although, I haven't searched around much thereafter. I am familiar with the content you've referred to above, so I think my inquiry is different. Funny thing is that I haven't bothered reading the paper I referenced. Perhaps I should give it a swirl.
 
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