Timeline for Bézout and products in algebraic groups
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2018 at 12:50 | history | edited | H A Helfgott | CC BY-SA 4.0 |
added 16 characters in body
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| Oct 15, 2018 at 11:31 | history | edited | YCor |
edited tags; edited tags
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| Oct 15, 2018 at 10:41 | history | edited | H A Helfgott | CC BY-SA 4.0 |
oops - had said deg instead of dim
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| Oct 15, 2018 at 10:40 | comment | added | H A Helfgott | Oh, I see - I said $\deg(V)+\deg(W)\geq \deg(G)$ when I meant $\dim(V)+\dim(W)\geq \dim(G)$. My bad. | |
| Oct 13, 2018 at 16:23 | comment | added | H A Helfgott | On the first comment by inkwell: well, sure, let us fix a faithful representation of $G$. On the second comment: there is something I am not seeing. For generic $V$ and $W$ of dimension $d\geq \deg(G)/2$ (say), surely, $V\cap a W$ be $0$-dimensional for $a$ generic? In that case, $\overline{V\cdot W^{-1}}$ should indeed be of dimension $\deg(G)$. | |
| Oct 13, 2018 at 7:36 | comment | added | inkspot | More substantively, I'm confused by the statement ''if $\deg(V)+\deg(W)\ge\deg(G)$, then, generically, $\overline{V⋅W}$ is the entire group". If $V,W$ are curves of high degree and $\dim G\ge 3$ this is false if $V.W$ denotes the image in $G$ of $V\times W$ under the multiplication morphism. | |
| Oct 13, 2018 at 7:29 | comment | added | inkspot | This starts with the choice of a faithful representation of $G$. Then ''degree'' will mean ''degree of the closure in $\mathbb P^{n^2}$. Requiring $\deg G=1$ will then require dividing by $\deg G$ so that degree is now a rational number. | |
| Oct 12, 2018 at 21:09 | comment | added | H A Helfgott | Well, I was thinking of everything as sitting inside $\mathbb{A}^{n^2}$, since we can see $G$ as a subgroup of $\GL_n$. At the same time, I agree that it would be helpful to have some definition of degree of subvarieties of $G$ that is in some sense relative to $G$ (in such a way that the degree of $G$ relative to itself is $1$). | |
| Oct 12, 2018 at 20:48 | comment | added | inkspot | Laurent Moret-Bailly's question is still pertinent. Into which affine or projective space are you embedding your varieties? "Degree" is not defined without choosing such an embedding. | |
| Oct 12, 2018 at 20:15 | comment | added | H A Helfgott | ... though I think Bézout is still true if you define the degree of a reducible variety to be the sum of the degrees of all irreducible components. Yup, see the Fulton reference in mathoverflow.net/questions/42127/… | |
| Oct 12, 2018 at 20:15 | comment | added | H A Helfgott | I just said that above! | |
| Oct 12, 2018 at 19:02 | comment | added | Laurent Moret-Bailly | What is the degree? | |
| Oct 12, 2018 at 16:24 | comment | added | YCor | [you can write italics/boldface in comments surrounding with star sign, resp double star ($*$blah$*$ yields blah and $**$blah$**$ yields blah)] | |
| Oct 12, 2018 at 16:12 | history | edited | H A Helfgott | CC BY-SA 4.0 |
correction taking into account YCor's comment below
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| Oct 12, 2018 at 16:07 | comment | added | H A Helfgott | What {\em is} the case is that $\dim(\overline{V\cdot W})$ has dimension $\dim(V)+\dim(W)-\dim(V\cap a W^{-1})$ for $a\in \overline{V\cdot W}$ generic (or for $a$ in the image of a generic point of $V\times W$ under the multiplication map, which is the same). | |
| Oct 12, 2018 at 16:05 | comment | added | H A Helfgott | Hm, you are right. For the rest of the question - let us go with "sum of the degrees of irreducible components of maximal dimension" as the definition of the degree of reducible variety. | |
| Oct 12, 2018 at 16:00 | comment | added | YCor | The claim that $\dim(\overline{V\cdot W})$ has dimension $\dim(V)+\dim(W)-\dim(V\cap W)$ is not true. For instance, typically when $V=W$ (say, for a general enough curve in the plane), we have $\dim(V\cdot V)=2\dim(V)$ (and not $=\dim(V)$). For the rest of the question, I'm not sure what you call degree. | |
| Oct 12, 2018 at 13:26 | history | asked | H A Helfgott | CC BY-SA 4.0 |