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Jordan curve theorem for cylinders
Allen: thanks for fixing the broken link!
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Jordan curve theorem for cylinders
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Blackbox Theorems
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Blackbox Theorems
Oh, ok. Do you want to declare the two answers merged for purpose of this voting people are doing? (since this is community wiki anyway)
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If graph is tree what can be said about its adjacency matrix ?
@Douglas: I think the question of what are the eigenvalues and eigenvectors of the adjacency matrix of a bipartite graph seems quite interesting. And thinking about it below (with your much appreciated assistance) led me to observe some things I hadn't known before about symmetric functions -- that you can take any of the usual bases for symmetric functions (power sum, elementary, Schur, etc.), look at the algebra generated by only the generators of even degree (e.g. kill the odd degree power sums as below) and get the ring of symmetric functions in the new variables $x_1^2, x_2^2, \dots $.
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If graph is tree what can be said about its adjacency matrix ?
Actually, you can show that the elementary symmetric functions $e_i$ for $i$ odd are all 0 by expressing them in terms of the power sum symmetric functions and seeing that each summand is 0. Then shows that the coefficients for odd powers of $x$ in your original $P(x)$ are 0, implying $P(x)$ is an even function. This is interesting to think about what happens when you take the symmetric functions and mod out by the elements of odd degree in a basis. Thanks!
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If graph is tree what can be said about its adjacency matrix ?
Thanks! I'm a little slow in understanding your comment though I'm afraid. I don't yet see how to deduce that $log P(x)$ is even. I'll keep thinking about it.
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If graph is tree what can be said about its adjacency matrix ?
Now I'm also curious about the general statement for real numbers $\lambda_1, \dots ,\lambda_n$ of whether knowing $\sum_{j=1}^n \lambda_j^i =0$ for all positive, odd integers $i$ implies that the nonzero reals come in pairs $\mu_j, \nu_j$ of equal magnitude and opposite sign. It looks like this might follow from a variant on the Vandermonde determinant being nonzero where the usual Vandermonde matrix entry $a_i^j$ is replaced by $\mu_i^j + \mu_i^{j-1}\nu_i + \mu_i^{j-2}\nu_i^2 + \cdots + \nu_i^j$.
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If graph is tree what can be said about its adjacency matrix ?
I see -- you have given a proof of my conjecture in the specific case of adjacency matrices of bipartite graphs (rather than working with the relations among real numbers), noting that the standard basis vectors are indexed by the nodes of the graph and that applying the adjacency matrix sends a node to the sum of standard basis vectors indexing its neighbors. That's a nice proof! Thanks!
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How often do people read the work that they cite?
I picked the Poincare Conjecture for a few reasons: (1) it has been carefully checked; (2) the probability is 0 that an error got past the experts, whereas I would find that error; (3) the statement looks to be very useful even to people who don't know its proof; (4) it would be an insane amount of work for most people, if they could do it at all, to check the proof. The five lemma is a different story, though I suppose I'm not sure who is originally responsible for developing/discovering it.
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