aglearner
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Registered User
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It is great to learn math on this website.
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May 20 |
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A “Riemannian” analogue of Kobayashi metric? Dear Curtis, thank you for this remark! Have I got you correctly, that you put $U$ inside $\partial H^3$. Do you know if someone studied the definition that I propose? |
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May 20 |
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Possible automorphism groups of a K3 surface The answer does not change if you remove "polarized" provided your K3 surface is projective. Indeed take any positive integral (1,1) form corresponding to a polarization of K3 and sum it over the action of the group. This will give you an invariant positive integral (1,1) form and hence an invariant polarisation. I don't know if non-algebraic K3 surfaces can have an automorphism of finite order (if this is possible the automorphism will preserve the volume form so that the quotient is a singular K3 again) |
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May 15 |
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A “Riemannian” analogue of Kobayashi metric? Misha, thanks for your comment I'll check the paper. |
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May 14 |
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Is it true that Nature promotes products? Thank you, this is a sharp answer! |
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May 14 |
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A “Riemannian” analogue of Kobayashi metric? deleted 1 characters in body |
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May 14 |
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Is it true that Nature promotes products? added 210 characters in body |
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May 14 |
asked | Is it true that Nature promotes products? |
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May 14 |
asked | A “Riemannian” analogue of Kobayashi metric? |
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Apr 25 |
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Conical divisor over a $\mathbb Q$-Cartier divisor. Dear Karl, yes this is very nice thank you for expanding your answer! |
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Apr 23 |
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Conical divisor over a $\mathbb Q$-Cartier divisor. Dear Karl, yes I would be grateful if you could give more details. |
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Apr 23 |
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Conical divisor over a $\mathbb Q$-Cartier divisor. Dear Karl, many thanks for you answer! Do I understand correctly that in the first formula for $O_Y(D_Y)$ the term on the right of the equality is the space of global sections of $O_Y(D_Y)$? |
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Apr 22 |
asked | Conical divisor over a $\mathbb Q$-Cartier divisor. |
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Apr 22 |
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$H^1(X,O_X)$, holomorphic $1$-forms, and $b_1(X)/2$ for normal $X$. Also, I would like to ask you if you can propose some reference where I could read about mixed Hodge structure (to get a more complete understanding of your answer). |
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Apr 22 |
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$H^1(X,O_X)$, holomorphic $1$-forms, and $b_1(X)/2$ for normal $X$. Dear Donu many thanks for your answer (great that you have not retired from here completely :) )! In fact in the situation I am dealing with $X$ is Moishezon (and not necessarily projective). Do I understand correctly that your argument still works in such a case? |
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Apr 22 |
asked | $H^1(X,O_X)$, holomorphic $1$-forms, and $b_1(X)/2$ for normal $X$. |
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Apr 5 |
awarded | ● Popular Question |
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Mar 30 |
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An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved. Thanks a lot, this is helpful! |
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Mar 13 |
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Injective morphism from curves to $\mathbb CP^2$ Vivek, huge thanks for this update! Now I am completely convinced in your statement. Your answer is really nice (the only reason I have not yet accepted it is that it seems to me now that the original question is an open problem...) |
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Mar 5 |
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Higher dimensional Bezout via Hilbert polynomials: a reference. added 238 characters in body; edited tags |
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Mar 5 |
awarded | ● Yearling |
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Mar 4 |
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Higher dimensional Bezout via Hilbert polynomials: a reference. Abdelmalek, thank you for these references! It looks to me indeed that Van der Waerden proves Bezout using only Nullstelensatz, this is very nice :) But it will take me some time of course, to understand the proof. |
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Mar 4 |
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Higher dimensional Bezout via Hilbert polynomials: a reference. Abdelmalek, thank you for your remark. To my shame I don't know what is the multidimensional resultant. Could you tell where can I read about this? I would like to learn more about the approach you propose. |
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Mar 4 |
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Higher dimensional Bezout via Hilbert polynomials: a reference. Sandor, I agree that claim 1 from your answer would suffice. Also, Manin's suggested proof of Bezout goes as follows: you calculate by induction the leading coefficient of the Hilbert polynomial of $A_r=k[x_0,....,x_n]/(f_1,...,f_r)$. When you make a step of induction and pass form the Hilbert polynomial of $A_r$ to the Hilbert polynomial of $A_{r+1}$ the easiest way to proceed would be to say that $f_{r+1}$ is not a zero divisor in $A_r$...Hartshorne proposes a more detailed analysis of what happens at this step (introducing multiplicities), but I don't see how to state all this with ease... |
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Mar 4 |
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Higher dimensional Bezout via Hilbert polynomials: a reference. Dear Sandor, thank you for this very detailed answer! This helps me to understand better the nature of regular sequences. This also indicates me I guess that I will need to be less ambitious in what to say to students :)... |
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Mar 3 |
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Higher dimensional Bezout via Hilbert polynomials: a reference. Jack I agree with what you say and I agree that this it is probably more important. But this is also harder to state (comparing to Bezout's theorem). |
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Mar 3 |
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Example of wall-crossing formulae? Liviu, I beg you pardon, it seems to me that the example you proposed tells us more or less how to solve quadratic equations with real coefficients. You can call this wall-crossing, but this sounds a little bit like giving a new name to a relatively old thing. For me this does not quite answer the question asked by Kim... I suspect that Kontsevich and Soibelman made some more substantial progress? (though I might be wrong...) |
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Mar 3 |
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Higher dimensional Bezout via Hilbert polynomials: a reference. Felipe, thanks. I am afraid that what you propose would not fit into my course... I have 4 more lectures to give and finished with Hilbert basis theorem last time. It will take me more than four lectures to go in details through section 1.7 of Hartshorne and this will kill my students (and myself I guess). So I am looking for some miraculous solution. If it does not exist I'll do just like Hasset and Manin (living this statement about regular sequence as an exercise :( ...) |
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Mar 3 |
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Higher dimensional Bezout via Hilbert polynomials: a reference. added 229 characters in body |
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Mar 3 |
asked | Higher dimensional Bezout via Hilbert polynomials: a reference. |
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Feb 28 |
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Injective morphism from curves to $\mathbb CP^2$ added 267 characters in body |
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Feb 28 |
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Injective morphism from an elliptic curve to $\mathbb CP^2$. Thank you Jeremy (I was in fact thinking of $x^{2n}+y^{2n}-z^{2n}=0$, but, of course I agree with your choice of tangent lines :) ) |
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Feb 27 |
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Injective morphism from an elliptic curve to $\mathbb CP^2$. Do I understand correctly that the same reasoning applies to the curve $x^{2n}+y^{2n}+z^{2n}=0$ for any $n$. Namely, lines $x=z$, $x=-z$ and $y=z$ have tangency of order $2n$ with such a curve and we can apply birational transformation of the type your proposed with respect to these three lines. I am asking this, since I had a secret hope that curves of (fixed) higher genus can not admit injections in $\mathbb CP^2$ of arbitrary high degree. But it looks now they can... So this tells me that it will be harder to answer in negative the original question 122645... |
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Feb 27 |
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Injective morphism from an elliptic curve to $\mathbb CP^2$. That is very cute. Just to be sure I get it, the image of this elliptic curve has exactly two cusps? |
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Feb 27 |
asked | Injective morphism from an elliptic curve to $\mathbb CP^2$. |
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Feb 26 |
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Injective morphism from curves to $\mathbb CP^2$ Vivek, thanks a lot! It will take me time to absorb your proof. |
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Feb 25 |
awarded | ● Nice Question |
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Feb 24 |
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Injective morphism from curves to $\mathbb CP^2$ added 205 characters in body; added 2 characters in body |
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Feb 24 |
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A classification of rational surfaces with effective $K$ Thank you Rita and sorry for doubting the answer I finally got it:) I will accept it. |
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Feb 24 |
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A classification of rational surfaces with effective $K$ Thank you for these details. I still miss something. Consider the double cover of $\hat{\mathbb P^2}$ in $B'$. Then the preimage of the exceptional divisor in $\hat{\mathbb P^2}$ is an ellipitic curve. I think, that there is a projection from $Y$ to this curve collapsing all the rational curves from the "free rational pencil". Am I wrong? (if not, clearly $Y$ is not rational) |
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Feb 24 |
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A classification of rational surfaces with effective $K$ Gianni, sorry my second comment is wrong indeed, I will delete it |
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Feb 24 |
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A classification of rational surfaces with effective $K$ Dear Rita, are you sure that your surface is rational and not ruled over an elliptic curve? In fact, could you please say what is "ordinary quadruple point"? (is this something like $x^4=y^4$?) Basically I am not sure that you get a pencil of lines on $Y$ -- you get one on $X$, but will they all pass through one point on $Y$? |
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Feb 24 |
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A classification of rational surfaces with effective $K$ Sorry Gianny, this does not quite make sense for me. You seem to claim that on any rational surface anticanonial divisor is big. This is unfortunately wrong, it need not be effective even. Just take a blow up of $\mathbb P^2$ in $10$ generic points. |
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Feb 24 |
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“Arithmetic genus” of a plane curve singularity. Thank you for this answer Jeremy! |
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Feb 24 |
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“Arithmetic genus” of a plane curve singularity. added 102 characters in body |
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Feb 24 |
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A classification of rational surfaces with effective $K$ the question is slightly specified |
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Feb 24 |
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“Arithmetic genus” of a plane curve singularity. Vivek, thank you this recommendation and as well for the explanation of what is the conductor. |
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Feb 23 |
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“Arithmetic genus” of a plane curve singularity. It seems to me that Milnor conjecture concerns only toric knots. |
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Feb 23 |
asked | “Arithmetic genus” of a plane curve singularity. |
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Feb 23 |
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An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved. deleted 58 characters in body |
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Feb 22 |
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Injective morphism from curves to $\mathbb CP^2$ added 90 characters in body |
