MBS - Revision history

Anton at 00:52, 2 December 2011

2011-12-02T00:52:02Z

← Older revision		Revision as of 16:52, 1 December 2011
Line 19:		Line 19:
	There are some obvious candidates for other cases, but they don't work so well. For example, consider the family $k[x,y](x^2,xy,y^2)$ over $k$. It feels natural to use relations like $X^2-sym(x\otimes x\otimes 1)$ (where $sym$ means $1/6$ of the symmetrization) or equivalently $X^2-sym(x\otimes 1\otimes 1)^2$. However, it looks like the discriminants you get this way turn out to be zero.		There are some obvious candidates for other cases, but they don't work so well. For example, consider the family $k[x,y](x^2,xy,y^2)$ over $k$. It feels natural to use relations like $X^2-sym(x\otimes x\otimes 1)$ (where $sym$ means $1/6$ of the symmetrization) or equivalently $X^2-sym(x\otimes 1\otimes 1)^2$. However, it looks like the discriminants you get this way turn out to be zero.

			=Chat with Matt [[User:Anton\|Anton]] 16:52, 1 December 2011 (PST)=

			Not clear if vanishing locus of the discriminent actually contains the complement of the principal component, but since the hilbert scheme is finite type, there does exist some ''power'' of the discriminent which contains the complement of the principal component. So to show that a given 0-scheme is smoothable, it is enough to show it is the "closed fiber" of a family where the discriminent has sufficiently large order of nilpotence.

	[[Category:Note]]		[[Category:Note]]

Anton at 02:20, 27 November 2011

2011-11-27T02:20:05Z

← Older revision		Revision as of 18:20, 26 November 2011
Line 7:		Line 7:

	The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation (at least when the characteristic is bigger than $n$). [[User:Anton\|Anton]] 21:06, 17 November 2011 (PST)		The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation (at least when the characteristic is bigger than $n$). [[User:Anton\|Anton]] 21:06, 17 November 2011 (PST)

			{{question\|show that various properties about Galois closure follow from this description}}

	=Chat with Matt [[User:Anton\|Anton]] 17:58, 26 November 2011 (PST)=		=Chat with Matt [[User:Anton\|Anton]] 17:58, 26 November 2011 (PST)=
	* Grothendieck-Deligne map can be though of as follows: if $E$ over $B$ is an $n$-dimensional algebra, then the Hilbert scheme of $n$ points in $E$ is trivial, but $Sym^n(E)$ is not. The Chow map from Hilb to Sym is a closed immersion which corresponds to the GD map.		* Grothendieck-Deligne map can be though of as follows: if $E$ over $B$ is an $n$-dimensional algebra, then the Hilbert scheme of $n$ points in $E$ is trivial, but $Sym^n(E)$ is not. The Chow map from Hilb to Sym is a closed immersion which corresponds to the GD map.
	*		* {{question\|is the GD map stable under arbitrary base change? it should be}}
			* Note that for any smoothable family $E$ over $B$, there is a ''nilpotent'' extension of the family which witnesses smoothability. As soon as the discriminant is non-zero at every point of $B$, $E$ must be smoothable.

			Idea: Suppose $E$ over $B$ smoothable. Perhaps there is a universal nilpotent thickening of the family which witnesses smoothability, and maybe this is given by the GD map. For example, consider $k[x]/x^2$ over $k$, then the GD map is from $TS^2 k[\epsilon]$ (different letter used for clarity in what follows). Then the family $k[x,\epsilon]/(x^2-\epsilon\otimes \epsilon)$ witnesses that the original family is smoothable.

			There are some obvious candidates for other cases, but they don't work so well. For example, consider the family $k[x,y](x^2,xy,y^2)$ over $k$. It feels natural to use relations like $X^2-sym(x\otimes x\otimes 1)$ (where $sym$ means $1/6$ of the symmetrization) or equivalently $X^2-sym(x\otimes 1\otimes 1)^2$. However, it looks like the discriminants you get this way turn out to be zero.


	[[Category:Note]]		[[Category:Note]]

Anton at 02:04, 27 November 2011

2011-11-27T02:04:51Z

← Older revision		Revision as of 18:04, 26 November 2011
Line 1:		Line 1:
	<protect>Anton</protect>		<protect>Anton,Matt</protect>

	=Hilbert component=		=Hilbert component=
Line 9:		Line 9:

	=Chat with Matt [[User:Anton\|Anton]] 17:58, 26 November 2011 (PST)=		=Chat with Matt [[User:Anton\|Anton]] 17:58, 26 November 2011 (PST)=
			* Grothendieck-Deligne map can be though of as follows: if $E$ over $B$ is an $n$-dimensional algebra, then the Hilbert scheme of $n$ points in $E$ is trivial, but $Sym^n(E)$ is not. The Chow map from Hilb to Sym is a closed immersion which corresponds to the GD map.
			*


	[[Category:Note]]		[[Category:Note]]

Anton at 01:58, 27 November 2011

2011-11-27T01:58:43Z

← Older revision		Revision as of 17:58, 26 November 2011
Line 7:		Line 7:

	The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation (at least when the characteristic is bigger than $n$). [[User:Anton\|Anton]] 21:06, 17 November 2011 (PST)		The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation (at least when the characteristic is bigger than $n$). [[User:Anton\|Anton]] 21:06, 17 November 2011 (PST)

			=Chat with Matt [[User:Anton\|Anton]] 17:58, 26 November 2011 (PST)=


	[[Category:Note]]		[[Category:Note]]

Anton at 17:18, 19 November 2011

2011-11-19T17:18:52Z

← Older revision		Revision as of 09:18, 19 November 2011
Line 1:		Line 1:
	<protect>Anton</protect>		<protect>Anton</protect>

			=Hilbert component=
	Suppose $E$ is an algebra over $B$, free of rank $n$ as a module. Then we have the Grothendieck-Deligne norm map $\sigma:TS^nE\to B$. This map is determined by two properties:		Suppose $E$ is an algebra over $B$, free of rank $n$ as a module. Then we have the Grothendieck-Deligne norm map $\sigma:TS^nE\to B$. This map is determined by two properties:
	# it is a $B$-algebra homomorphism,		# it is a $B$-algebra homomorphism,

Anton: moved Hilbert component to MBS

2011-11-19T17:17:45Z

moved Hilbert component to MBS

← Older revision	Revision as of 09:17, 19 November 2011
(No difference)

Anton at 17:16, 19 November 2011

2011-11-19T17:16:46Z

← Older revision		Revision as of 09:16, 19 November 2011
Line 1:		Line 1:
			<protect>Anton</protect>

	Suppose $E$ is an algebra over $B$, free of rank $n$ as a module. Then we have the Grothendieck-Deligne norm map $\sigma:TS^nE\to B$. This map is determined by two properties:		Suppose $E$ is an algebra over $B$, free of rank $n$ as a module. Then we have the Grothendieck-Deligne norm map $\sigma:TS^nE\to B$. This map is determined by two properties:
	# it is a $B$-algebra homomorphism,		# it is a $B$-algebra homomorphism,

Anton at 03:30, 19 November 2011

2011-11-19T03:30:34Z

← Older revision		Revision as of 19:30, 18 November 2011
Line 3:		Line 3:
	# $\sigma(x\otimes x\otimes\cdots\otimes x)=N_{E/B}(x)$ for any $x\in E$.		# $\sigma(x\otimes x\otimes\cdots\otimes x)=N_{E/B}(x)$ for any $x\in E$.

	The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation. [[User:Anton\|Anton]] 21:06, 17 November 2011 (PST)		The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation (at least when the characteristic is bigger than $n$). [[User:Anton\|Anton]] 21:06, 17 November 2011 (PST)

	[[Category:Note]]		[[Category:Note]]

Anton at 03:30, 19 November 2011

2011-11-19T03:30:01Z

← Older revision		Revision as of 19:30, 18 November 2011
Line 3:		Line 3:
	# $\sigma(x\otimes x\otimes\cdots\otimes x)=N_{E/B}(x)$ for any $x\in E$.		# $\sigma(x\otimes x\otimes\cdots\otimes x)=N_{E/B}(x)$ for any $x\in E$.

	The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to 1. This demonstrates that $\overline E$ always contains a copy of the sign representation. [[User:Anton\|Anton]] 21:06, 17 November 2011 (PST)		The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to $n!$. This demonstrates that $\overline E$ always contains a copy of the sign representation. [[User:Anton\|Anton]] 21:06, 17 November 2011 (PST)

	[[Category:Note]]		[[Category:Note]]

Anton: Created page with "Suppose $E$ is an algebra over $B$, free of rank $n$ as a module. Then we have the Grothendieck-Deligne norm map $\sigma:TS^nE\to B$. This map is determined by two properties: # ..."

2011-11-18T05:06:56Z

Created page with "Suppose $E$ is an algebra over $B$, free of rank $n$ as a module. Then we have the Grothendieck-Deligne norm map $\sigma:TS^nE\to B$. This map is determined by two properties: # ..."

New page

Suppose $E$ is an algebra over $B$, free of rank $n$ as a module. Then we have the Grothendieck-Deligne norm map $\sigma:TS^nE\to B$. This map is determined by two properties:
# it is a $B$-algebra homomorphism,
# $\sigma(x\otimes x\otimes\cdots\otimes x)=N_{E/B}(x)$ for any $x\in E$.

The kernel of $\sigma$ is the ideal generated by the norm relations of the Galois ideal. The Galois ideal is generated by these same relations, so the Galois ideal is induced by the kernel of $\sigma$. This shows that the Galois closure of $E$ is $T^nE\otimes_{TS^nE}B$. Considering the $TS^nE$-linear map $T^nE\to TS^nE$ given by $\alpha(-)/\alpha(x)$, we get a $B$-linear map from $\overline E$ to $B$ which sends $\alpha(x)$ to 1. This demonstrates that $\overline E$ always contains a copy of the sign representation. [[User:Anton|Anton]] 21:06, 17 November 2011 (PST)

[[Category:Note]]