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1d |
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Why is the inclusion of the tensor product of the duals into the dual of the tensor product not an isomorphism? added 1 characters in body |
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1d |
awarded | Nice Answer |
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1d |
awarded | Constituent |
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2d |
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Degree of a Cartier Divisor under pullback Dear Dedalus, Thanks for the reply. One way to think about it is that degree is a concept for a projective variety. The only time it intrinsic is for points (when it just counts the number of points). In the setting of a codim'n one point on divisors in a higher dimensional varieties, the multiplicity counts the "number of times" the particular irred. component corresponding to that point appears in the divisor. Regards, |
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2d |
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Degree of a Cartier Divisor under pullback Dear Dedalus, I think you need to assume that $Y$ is of dimension one, otherwise the multiplicity will be infinite if $y \in D$ is a non-generic point. (And, without fixing a projective embedding, degree is only meaningfully defined for divisors in curves.) Regards, |
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May 17 |
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Prove that $\log X < X$ for all $X > 0$ added 14 characters in body |
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May 17 |
answered | Prove that $\log X < X$ for all $X > 0$ |
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May 17 |
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Prove that $\log X < X$ for all $X > 0$ Dear A Walker, The OP was also (perhaps principally) interested in the case of $\log$ to the base $2$, rather than natural log. Regards, |
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May 17 |
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Computing Chern Classes of Tautological Line Bundles Dear Olivier, I agree, based on the analogy with complex projective case. Cheers, |
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May 17 |
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what is the relation of smooth compact supported funtions and real analytic function? Dear Sara, I know this is not directly what you asked about, but there is a more subtle relationship: the Fourier transform of a test function will be an analytic function (in fact an entire function of a complex variable). Regards, |
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May 17 |
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Computing Chern Classes of Tautological Line Bundles Dear Olivier, This looks good to me; I was also a bit unsure about the sign. Part of the issue perhaps is that one has to figure out how $P(\mathbb H^2)$ is oriented in the first place. I guess the quaternionic situation is something like the complex situation, where one gets a canonical orientation (?) . Regards, |
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May 17 |
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Computing Chern Classes of Tautological Line Bundles @OlivierBégassat: Dear Olivier, What you write would be consisent with $c_1(V) = 0$ (which I think can also be thought of as being caused by the fact that an $\mathbb H$-line bundle, when though of as a complex vector bundle, is actually an $SU(2)$-bundle, and so its determinant is trivial). But I don't think it would cause $c_2$ to vanish, since this would presumably be exactly be the degree $4$ class you're thinking of. Cheers, |
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May 17 |
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Computing Chern Classes of Tautological Line Bundles Dear mathdragon, I would guess that $c_2$ is equal to the cohomology class of $P(\mathbb H)$ inside $P(\mathbb H^2)$, or perhaps its negative (I'm unsure about the conventions being used right now), and I might be able to prove it, starting from the definition I'm most comfortable with, which is definitely not the splitting principle. Regards, |
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May 17 |
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Computing Chern Classes of Tautological Line Bundles @OlivierBégassat: Dear Olivier, I was bit confused about this at first, but isn't it the case that this is canonically a bundle of $1$-dimensional $\mathbb H$-modules, and by fixing a copy of $\mathbb C$ inside $\mathbb H$ (which is somewhat canonical, since they're all conjugate in $\mathbb H$) we can regard it as a bundle of rank $2$ complex vector spaces? Regards, |
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May 17 |
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Computing Chern Classes of Tautological Line Bundles Dear mathdragon, One more thing: if the fibres are $\mathcal H$-lines in $\mathcal H$ (which I would guess they are, since this is the usual definition of the tautological bundle), then aren't they $2$-dimensional over $\mathbb C$, making $V$ a rank $2$ complex vector bundle? (You write line bundle in your post, which is why I am asking.) Regards, |
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May 17 |
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Computing Chern Classes of Tautological Line Bundles Dear mathdragon, Just to check, $\mathbb H$ denotes the Hamilton quaternions? Regards, |
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May 17 |
answered | Quotient rings of Gaussian integers $\mathbb{Z}[i]/(2)$, $\mathbb{Z}[i]/(3)$, $\mathbb{Z}[i]/(5)$, |
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May 17 |
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For $I,J$ ideals, show that $IJ\subseteq (I\cap J)(I+J)$ Note that this question is closely related. |
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May 17 |
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Closed subset of an affine variety… is it affine? @Abramo: Dear Abramo, Actually, looking again at Georges's answer, there is another comment to make: namely, he has already observed that $X_1 \times a_2$ is a variety (i.e. is irreducible), and the only issue now is whether or not it is an affine variety --- this is where he uses the fact that it is closed in something affine. So another sense in which you can interpret his argument is that he is saying that an irreducible closed subset of an affine variety is again an affine variety (which is again essentially a matter of definitions). Regards, |
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May 17 |
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Closed subset of an affine variety… is it affine? ... about the irreducibility, or non-irreducibility, of the set in question. He is simply saying that (essentially by definition) a closed subset of an affine algebraic set is again an affine algebraic set. Regards, |
