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Jul
28 |
comment |
Transcendental distance sets
The same sort of argument can be used if you're avoiding any countable collection of algebraic relations. |
Jul
28 |
comment |
Transcendental distance sets
You can use Zorn's lemma and measure theory. Suppose $S$ is a maximal subset of $\mathbb{R}^n$ with no algebraic volumes - one exists by Zorn's lemma. Let $D$ be the set of all points $z$ of $\mathbb{R}^n$ such that $z$ forms a simplex of algebraic volume together with some $n$ points of $S$. Since the set of $n$-tuples of points of $S$ is countable, and the set of algebraic numbers is countable, you see that $D$ is a union of a countable collection of hyperplanes, thus has measure zero. This means that there is a point in the complement to $D\cup S$, so $S$ is not maximal - contradiction! |
Jun
26 |
awarded | Yearling |
May
10 |
comment |
When is the Hom-scheme connected?
Whether or not you assume commutativity, you won't in general get connectivity. Take for example maps from the n'th neighborhood of $0$ in $\mathbb{A}^1$ into the n'th neighborhood of $0$ in $Spec(k[x,y]/(xy)$. These mapping spaces are in general at least as complicated as mapping spaces of fixed degree between projective varieties: if $A$ is functions on a neighborhood of the affine cone of a projective variety $X$ and $B$ is functions in a neighborhood of 0 in a veronese twisted affine cone of $Y$ then $Map(B,A)$ goes to $Map(X,Y)$ surjectively by taking blowups of tangent cones. |
May
10 |
answered | Fundamental group of the complement homogeneous variety in $\mathbb{C}P^{n-1}$ |
Apr
27 |
answered | Is there a nonzero sheaf with all cohomologies vanish? |
Apr
27 |
asked | Twisting stable maps to C* equivariant space by a line bundle |
Apr
5 |
answered | On a theorem of Kazhdan |
Mar
25 |
comment |
On a theorem of Kazhdan
Ah sorry, thought you were twisting by a character of the Levi (which would not affect the dimension... your current question makes a lot more sense). |
Mar
24 |
revised |
On a theorem of Kazhdan
minor correction |
Mar
24 |
awarded | Informed |
Mar
24 |
revised |
On a theorem of Kazhdan
minor edit |
Mar
24 |
answered | On a theorem of Kazhdan |
Mar
18 |
awarded | Popular Question |
Feb
11 |
awarded | Self-Learner |
Jan
12 |
revised |
Twisting of the power functor
minor edit: added the word "perfect" in a couple places |
Jan
12 |
comment |
Twisting of the power functor
Hi David, I mean the category of perfect modules over $T(k)$, which I think it's enough to consider as a DG algebra over $k$. (Correct me if this doesn't make sense). |
Jan
11 |
revised |
Twisting of the power functor
edited typo |
Jan
11 |
asked | Twisting of the power functor |
Dec
1 |
accepted | What is the coefficient ring of algebraic K theory of the discrete $\mathbb{C}$? |