Feb
1 |
comment |
Definition of E-infinity operad
Crosspost: math.stackexchange.com/questions/1636588/… |
Feb
1 |
comment |
Is there a nice choice-free argument to count the number of sublattices?
@Simon: here is an even simpler example: the number of points in an $n$-dimensional vector space over $\mathbb{F}_q$ is $q^n$. How do you prove this without picking a basis? For that matter, how do you define "$n$-dimensional" without picking a basis? (In both cases you can again instead pick a complete flag, but I don't know how to pick less than this; said another way, I don't know how to make the argument equivariant with respect to any group larger than a Borel subgroup of $GL_n(\mathbb{F}_q)$.) |
Feb
1 |
revised |
Is there a nice choice-free argument to count the number of sublattices?
deleted 29 characters in body |
Feb
1 |
revised |
Is there a nice choice-free argument to count the number of sublattices?
added 115 characters in body |
Feb
1 |
answered | Is there a nice choice-free argument to count the number of sublattices? |
Feb
1 |
comment |
Is there a nice choice-free argument to count the number of sublattices?
@Tom: matrices in $GL_2(\mathbb{Z})$ can only have determinant $\pm 1$. |
Feb
1 |
comment |
Representation theory and associated bundles
Versions of this statement in the algebraic setting can be deduced from versions of Tannaka duality. |
Jan
31 |
comment |
If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$?
@Area: no, but they're related. For the correct statement look up the Hodge decomposition. On a compact Riemann surface $H^0(X, \mathcal{O}_X) \cong \mathbb{C}$ and $H^1(X, \mathcal{O}_X) \cong \mathbb{C}^g$ (not $2g$), and all higher cohomology vanishes. |
Jan
31 |
comment |
If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$?
@Area: the sheaf of continuous functions to $\mathbb{C}$ is fine, so for reasonable $X$ that means it's acyclic (that is, all higher cohomology vanishes). Your argument doesn't seem to have enough resolution to distinguish between the sheaf of continuous functions and the sheaf of holomorphic functions. |
Jan
31 |
comment |
Galois cohomology of a non-abelian group over a function field
Do you mean $G(\overline{F})$? |
Jan
31 |
comment |
If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$?
...exact sequence in cohomology part of which goes $\dots \to H^0(X, \mathbb{C}) \to H^0(X, \mathbb{C}^{\times}) \to H^1(X, \mathbb{Z}) \to 0$. This shows that $H^1(X, \mathbb{Z})$ (once you believe that sheaf cohomology of the constant sheaf $\mathbb{Z}$ agrees with singular cohomology; let me assume $X$ is a reasonable space so this is true) is the quotient of the group of continuous functions $X \to \mathbb{C}^{\times}$ by the subgroup of exponentials of continuous functions $X \to \mathbb{C}$; this turns out to be the connected component of the identity (exercise). |
Jan
31 |
comment |
If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$?
@Area: on $C(X, \mathbb{C})$ the spectral radius and sup norm coincide. The group structure is induced from that of $S^1$. What I have in mind in general is that cohomology is represented by Eilenberg-MacLane spaces, and $S^1$ is the Eilenberg-MacLane space $B \mathbb{Z}$. In the particular case of $S^1$ you can also argue using the exponential sheaf sequence, as follows. There is a short exact sequence $0 \to \mathbb{Z} \to \mathbb{C} \to \mathbb{C}^{\times} \to 0$ of sheaves on $X$ (here each object stands for the sheaf of continuous functions into that object), which induces a long... |
Jan
31 |
answered | How do you rigidify a Bousfield localization? |
Jan
31 |
comment |
If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$?
@Area: the group of connected components of the group of units is the group of homotopy classes of maps $X \to \mathbb{C}^{\times}$. The latter space is homotopy equivalent to $S^1$, so equivalently we're talking about the group of homotopy classes of maps $X \to S^1$. And this is of course precisely $H^1(X, \mathbb{Z})$. |
Jan
31 |
comment |
If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$?
@Area: I don't really understand your question; it seems to be at least two different questions at once, which seem worth separating. First, it sounds like you're worried that I haven't distinguished higher genus surfaces, but I have: this is covered in Edit #1. Second, there are lots of interesting things to say about, say, principal $G$-bundles and $G$-local systems on a surface but it's not obvious that this information is easily accessible starting from the ring of functions (although there are things to say here). |
Jan
31 |
comment |
How should $A^α$ be defined for real $α ∈ [0,∞)$ and $A\in M_n(\mathbb C)$?
@Yemon: sure, I'm aware that $A$ admits a logarithm, but that's very different from saying that the logarithm $\log A$ is well-defined. |
Jan
31 |
comment |
How should $A^α$ be defined for real $α ∈ [0,∞)$ and $A\in M_n(\mathbb C)$?
What's your definition of $\log A$ if, say, $A$ has complex eigenvalues of absolute value greater than $1$? I think the only good answer here is for the case that $A$ is positive semidefinite where we can appeal to the continuous functional calculus. |
Jan
30 |
revised |
Uniqueness of the fusion ring for simple finite group
deleted 3 characters in body |
Jan
30 |
comment |
Uniqueness of the fusion ring for simple finite group
@David: don't you also need to know the converse? The converse isn't clear to me a priori. |
Jan
29 |
comment |
Can $C^*$-algebra of continuous functions on $R^n$ ($S^n$) be characterized alternatively?
Right. I had in mind the commutative case. |