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answered  A reference about Grassmannian over finite fields 
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“Dimension” of ideals in $F_q[x]/\langle x^n1\rangle$?
It's very misleading to call this "polynomials of degree less than $n$." That defines a vector space, not a ring: you've chosen a very particular ring structure on it by choosing what $x^n$ ends up being, and you would get a different ring structure on it by making a different choice. It's this choice that's responsible for what different ideals look like. 
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awarded  Good Answer 
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awarded  Nice Answer 
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Set of Special Unitary Matrices that are dense in SU(4) and obey certain relations
Correcting my nowdeleted answer, without the third relation this is the Artin presentation of the braid group $B_{N+1}$. The third relation describes some quotient of this group; maybe it's possible to say explicitly what this quotient is. 
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The fibration map $Diff(M) \rightarrow Emb(N,M)$
I don't understand what it means to ask whether a fibration exists over a map. Are you asking whether the map is a fibration? 
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answered  What (fun) results in graph theory should undergraduates learn? 
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Definition of Einfinity operad
Crosspost: math.stackexchange.com/questions/1636588/… 
Feb
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Is there a nice choicefree 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 $GL_n(\mathbb{F}_q)$.) 
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Is there a nice choicefree argument to count the number of sublattices?
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Is there a nice choicefree argument to count the number of sublattices?
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answered  Is there a nice choicefree argument to count the number of sublattices? 
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Is there a nice choicefree argument to count the number of sublattices?
@Tom: matrices in $GL_2(\mathbb{Z})$ can only have determinant $\pm 1$. 
Feb
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Representation theory and associated bundles
Versions of this statement in the algebraic setting can be deduced from versions of Tannaka duality. 
Jan
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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
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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
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Galois cohomology of a nonabelian group over a function field
Do you mean $G(\overline{F})$? 
Jan
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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
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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 EilenbergMacLane spaces, and $S^1$ is the EilenbergMacLane 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? 