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1d
comment Number of Plücker relations for a Grassmannian
I would use the Jacobian criterion math.stackexchange.com/questions/1258530/…
1d
comment Alternate proofs of Hilberts Basis Theorem
In the absence of dependent choice, a good definition is often "For any set $(I_a)_{a \in A}$ of ideals, there is an index $a$ such that, for all $b \in A$, the ideal $I_b$ does not properly contain $I_a$." I don't know how this plays with constructivity, though.
2d
answered Number of Plücker relations for a Grassmannian
Feb
5
awarded  Electorate
Feb
5
answered Nice proofs of the Poincaré–Birkhoff–Witt theorem
Feb
4
comment Intuition/idea behind a proof of the splitting principle?
Good point. I don't have an immediate intuition for that but I'll think about it
Feb
4
comment The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
Suggestion: Draw in the lines at $x=\tau y$ and $y = \tau x$, with $\tau = (1+\sqrt{5})/2$. I think that there are $10$ "wedges" which are permuted cyclically (the other boundaries are at $x=0$, $y=0$ and $x=-y$, which are already drawn.
Feb
4
answered Intuition/idea behind a proof of the splitting principle?
Feb
2
comment The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
Hmmm, what I wrote isn't exactly right. The orbit through $(8,27)$ crosses $x=8$. But I still think there is something here.
Feb
2
comment The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
Thanks for making this image! I notice that no orbits cross the lines $x=3$, $x=8$ or $x=21$. Can we prove this (and its generalization to other odd Fibonaccis)? If so, this fact coupled with its rotations under the $5$ fold symmetry proves the result.
Jan
29
comment Quotients of curves of genus $4$ by a free $\mathbb{Z}/ 3 \mathbb{Z}$-action
Of course, this invariant cubic is not unique; I am just saying there is one, well defined modulo the invariants in $H^0(\mathbb{P}^3, \mathcal{O}(1)) Q$
Jan
29
comment Quotients of curves of genus $4$ by a free $\mathbb{Z}/ 3 \mathbb{Z}$-action
Letting $Q$ be the invariant quadric, and $C$ the space of cubics, we have a short exact sequence $0 \to H^0(\mathbb{P}^3, \mathcal{O}(1)) Q \to C \to \mathbb{C} \to 0$ of $\mathbb{Z}/3$ representations, where $\mathbb{Z}/3$ acts trivially on the final $\mathbb{C}$. Such a sequence must split, the image of the splitting $\mathbb{C} \to C$ is an invariant cubic. (More explicitly, take any cubic $f$ not in $H^0(\mathbb{P}^3, \mathcal{O}(1)) Q$, then $(1/3) (f+\sigma f + \sigma^2 f)$ is an invariant cubic.)
Jan
29
awarded  Nice Question
Jan
29
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 $A$?
Choose $N$ large enough that $|w|/N < \pi$. Then $\exp(w/N)$ is disjoint from the negative real line, so $a \exp(w/N) + (1-a)$ is a unit for all $a \in [0,1]$. Thus $a v \exp((k+1) w/N) + (1-a) v \exp(k w/N)$ is a unit for all such $a$, and we can take a piecewise linear path through the points $v \exp(k w/N)$ for $0 \leq k \leq N$.
Jan
29
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 $A$?
Oh, sorry. I am confused by what you mean by "straight line". You mean literally straight in the vector space. That sounds plausible but I need to think a bit about whether it is actually true.
Jan
29
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 $A$?
@QiaochuYuan Yes, you are right. Obviously, if two units are connected by a line of units, they induce the same map $H_1(X) \to H_1(\mathbb{C}^{\ast})$. Conversely, if $u$ and $v$ are maps $X \to \mathbb{C}^{\ast}$ inducing the same map on homology, then $u v^{-1}$ has a continuous logarithm $w$, and then $v \exp(a w)$ is a continuous path from one to the other, for $a \in [0,1]$.
Jan
27
answered Hodge map and the Cohomology Ring of a Riemannian Manifold
Jan
24
comment the linear span of all matrix coefficients is $C(G,\mathbb{C})$ where $G$ is a finite group
I agree that this comes down to how pedantic you want to be. I think a maximal pedant would say that the proof is already complete if you delete all lines between "If the sum is zero...Hence the sum is nonzero.". It is clear that the $0$-function is in the space, so the sum is a vector space; the question is whether it is the vector space whose sole element is $0$. If you work with the assumption that it is $0$-vector space, then every step of the proof is correct until, at the end, you deduce that $\oplus V_{R_a} = \mathbb{C}^G$ (and, in particular, is not the $0$-vector space.)
Jan
23
awarded  Nice Answer
Jan
22
comment Curve with given Frobenius polynomial
You can still write the characteristic polynomial that way, just order the eigenvalues as $(\sqrt{p}, - \sqrt{p}, - \sqrt{p}, \sqrt{p})$.