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olivier dot begassat dot cours at gmail dot com


Jan
20
comment $\alpha \in \mathbb{C}$ with $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 2^k$ but $\alpha$ inconstructible
But $\zeta_6\cdot\sqrt{2}$ manifestly is constructible with a straight edge and a compass.
Jan
20
comment $S^1$ a p-local complex?
What about the usual reduced homology? $H_*(S^1)\simeq\Bbb Z[1]$ is a copy of the integers in degree $1$.
Jan
20
comment Laplacian and Hodge Laplacian
The Sobolev space of functions is just the Sobolev space of sections of the trivial line bundle.
Jan
20
comment Laplacian and Hodge Laplacian
I take it the Laplacian you talk about in the beginning is defined for real or complex valued functions, but you can define a Sobolev space of $p$-forms, and more generally, Sobolev spaces of sections of vector bundles.
Jan
20
comment Orbits of left-multiplication from $\mathrm{PSL}_2(\mathbb Z)$ on $\mathbb Z^{2\times 2}$
$SL_2$ acts on the left, not $PSL_2$.
Jan
19
comment Prove: $\int_0^1 \frac{\ln x }{x-1} d x=\sum_1^\infty \frac{1}{n^2}$
In case anybody reads this answer, I think it is necessary to say something like for all $u\in(0,1)$ and for all $n\in\Bbb N$, $$\left|1+\frac12u+\frac13u^2+\cdots\frac1nu^{n-1}\right|\leq-\frac{\ln(1-u)}u=g‌​(u)$$ so that, since $g$ is integrable on $(0,1)$, one can apply dominated convergence, and swap the sum with the integral.
Jan
19
comment Prove $F^2_{n+1} - F_nF_{n+2} = (-1)^n$
AlexR is correct!
Jan
19
comment Quotient space of $S^n$ and the projective plane
I doubt it: I don't thinkthe mapping cylinder is orientable when $n$ is even.
Jan
19
comment Mapping torus with homotopic homeomorphisms
@YonKim corrected the formulas.
Jan
19
comment Ext groups due to Yoneda: why is this class zero
Are you sure those modules are projective $\Bbb K[x]$-modules?
Jan
19
comment Is $\text{Ind}_{Z(G)}^G \rho$ irreducible or not for nonabelian group $G$?
It seems my first question is answered by the character formula for $\mathrm{Ind}\,W$, and this shows that the spanning set is indeed a basis, and the following comment shows that it splits, as a $Z$ module, into $[G:Z]$ copies of $W$. Ok, everything resolves itself : ) +1
Jan
19
comment Is $\text{Ind}_{Z(G)}^G \rho$ irreducible or not for nonabelian group $G$?
Also, if I'm not mistaken, if you consider $W=\Bbb C.w$, then it seems more correct to argue as follows: $$z\cdot(g\otimes w)=(zg)\otimes w=(gz)\otimes w=g\otimes (zw)=\rho(z)\cdot(g\otimes w)$$ so that $g\otimes w$, provided it is nonzero, spans a $Z$-submodule isomorphic to $W$ (they have the same character). Does that seem fair?
Jan
19
comment Is $\text{Ind}_{Z(G)}^G \rho$ irreducible or not for nonabelian group $G$?
How do you know that $\mathrm{Ind} W$ has dimension $|G|/|Z|$. I can see that there is a spanning set of this cardinality, but I don't see why it is necessarily a basis. Is it obvious?
Jan
19
comment Is $\text{Ind}_{Z(G)}^G \rho$ irreducible or not for nonabelian group $G$?
Complex irreducible representations of abelian groups are one dimensional... So $n=1$.
Jan
19
comment Mapping torus with homotopic homeomorphisms
Ok, there are some mistakes in the formulas, I'll get to them tomotopy.
Jan
19
comment Mapping torus with homotopic homeomorphisms
I don't know to be honest, but it seems like a good condition to impose.
Jan
19
comment Mapping torus with homotopic homeomorphisms
I mean a homotopy $H:X\times[0,1]\to X$ such that for all $t$, $H_t$ is a homeomorphism of $X$. In the case of the identity and the antipodal map of the sphere, you can take the homotopy $H(z,t)=e^{it\pi}z$, which is a homotopy through homeomorphisms.
Jan
19
comment Mapping torus with homotopic homeomorphisms
They are homotopic through homeomorphisms though. This is more restrictive than simply being homotopic in general. Do you see why the example I gave in my previous comment is a counter example?
Jan
19
comment Mapping torus with homotopic homeomorphisms
Do you mean simply homotopic or homotopic through homeomorphisms? In the first case the answer is no, take for instance $X=[0,1]$ and $f=\mathrm{id}$ and $g=1-\mathrm{id}$.
Jan
19
comment Quotient of homology groups
In general, $H_*(A)$ isn't included in $H_*(X)$, so the quotient doesn't make complete sense, although one could interpret it as the quotient of $H_*(X)$ by the image under $i_*$ of $H_*(A)$. Then you can use the long exact sequence to identify this with the image of $H_*(X)$ in $H_*(X,A)$.