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olivier dot begassat dot cours at gmail dot com


Jun
23
comment Notation for a vector space: $(\mathbb{C}^\infty)^{\otimes L}$
What do you mean by "countably or uncountably infinite"?
Jun
23
comment Notation for a vector space: $(\mathbb{C}^\infty)^{\otimes L}$
$\Bbb C^{\infty}$ is $$\bigoplus_{n\in\Bbb N}\Bbb C\,,$$ in other words, it's a complex vector space with numerable basis. You can describe it as the space of complex sequences that are eventually $0$.
Jun
23
comment Topological Vector Space: $\dim V=n\implies V\cong\mathbb{K}^n$
@user126154 $\Bbb R$ with the discrete topology isn't a TVS over $\Bbb R$ with its usual topology.
Jun
23
comment Topological Vector Space: $\dim V=n\implies V\cong\mathbb{K}^n$
I don't recall the proof in detail (it can be found in one of Rudin's books), but I remember that compactness is a key ingredient. Rudin requires $V$ to be Hausdorff. Then the image of the unit sphere (or ball) under $\phi$ is compact.
Jun
22
comment Homeomorphism proof
In the plane you can find simple homeomorphisms: first contract $R^2$ onto $R^*_+\times R$ via $(x,y)\mapsto(e^x,y)$, and then take the square by identifying the real plane with the complex numbers: $z\mapsto z^2$. This will give you a homeomorphism $R^2\to R^2\setminus R_-\times 0$ which you can then rotate to where you want it.
Jun
22
comment Trying to understand formula for counting regions of hyperplane arrangements in $\mathbb{R}^2$
@muffel for the first point: I'm saying that if you have (at most) $n-1$ points on a line, then those $n-1$ points divide the line into (at most) $n$ regions. For the second point, $$M_n\leq M_{n-1}+n\leq M_{n-2}+n-1+n\leq M_{n-3}+n-2+n-1+n\leq\cdots\leq M_1+2+3+\cdots+n-2+n-1+n$$
Jun
22
answered Trying to understand formula for counting regions of hyperplane arrangements in $\mathbb{R}^2$
Jun
22
revised $\exp(\ln(x))=x$ and $\ln(\exp(y))=y$.
Tried to clarify the question and the motivation behind it.
Jun
22
comment What kind of object is the kernel of a ring homomorphism?
What if you workrd with non unital rings? The zero ring is a zero object.
Jun
22
comment How to find volume of a sphere
Did you try to google "volume of a shpere" or "formula for the volume of a sphere"?
Jun
22
comment How to find volume of a sphere
Do you know the formula that gives the volume of a sphere given its radius $R$?
Jun
22
comment When does injectivity imply surjectivity?
$\exp{}{}{}{}{}$
Jun
22
comment Can we simultaneously realize arbitrary homotopy groups and arbitrary homology groups?
You don't consider the homotopy/homology groups of degree $>n$? If so $X=K(G_n,n)$ will do. Also only $G_1, H_1\sim G_1^{\mathrm{ab}}, H_2,\dots$ are required for your question in its current state.
Jun
22
comment Symmetry of Grassmanians
@YuriVyatkin thanks for adding a better drawing.
Jun
22
revised Symmetry of Grassmanians
corrected the argument
Jun
22
answered Symmetry of Grassmanians
Jun
21
comment Avoiding vertical vectors in tangent spaces.
Something fishy is going on: if we consider the differential $di$ induces (by virtue of $i$ being an embedding) a map $\tilde{di}$ between the projective bundles, then $\tilde{di}$ sweeps out an $2n-1$ dimensional submanifold of the $2(m+1)-1$-dimensional projective bundle $P(T(M\times[0,1]))$, and we want it to miss an $m+1+1-1$-dimensional submanifold. Now $(2n-1)+(m+1+1-1)=2n+m<2m+1=(2m+2-1)$. So it should work fine even when $2n=m$.
Jun
21
comment Avoiding vertical vectors in tangent spaces.
We most certainly won't be able to fix $dj\neq\iota$, but I don't think that's a problem.
Jun
21
comment Avoiding vertical vectors in tangent spaces.
I think $j$ thus defined will generically work, but that's just my guess. I'm still hung up on the dimensional argument though...
Jun
21
comment Avoiding vertical vectors in tangent spaces.
That's the point though: suppose $\iota$ is a perturbation of $di$, then you can define its "projection" $j=N\stackrel{0}{\hookrightarrow}TN\xrightarrow{\iota}T(M\times[0,1]) \xrightarrow{\pi}M\times[0,1]$, which will remain an embedding as long as $\iota$ is close to $di$, but you won't have $dj=\iota$. The problem is then that $dj$ might hit the vertical distribution, although I would venture to think generically it won't.