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olivier dot begassat dot cours at gmail dot com


Jun
27
comment Topologies on n-manifolds
Do you mean wether it is useful of even necessary to study general topology before learning about manifolds? I don't think so, at least if you are well acquainted with calculus and euclidean space. But you should know a little about compactness, and probably about covering space theory at some point.
Jun
27
comment Exact Sequences of R-Modules
This question is missing a lot of context. What are the maps? What is $E$?
Jun
26
revised Interpretation of $p$-forms
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Jun
26
comment Existence of $p \times p $ matrices $A$ and $B$ over the field $\mathbb F_p$, $p$ prime, such that $AB-BA=I$.
A related question math.stackexchange.com/questions/125219/… The answer is YES, but I don't know the proof, and it's not straightforward (at least the original proof isn't).
Jun
26
revised Interpretation of $p$-forms
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Jun
26
answered Interpretation of $p$-forms
Jun
25
comment Canonical orientation of a complex manifold
The orientation is different for even $n$. The second choice may have the added advantage that the transition matrices are bloc matrices with blocs of the form $$\begin{pmatrix} a&-b\\b&a\end{pmatrix}$$ and so identifies nicely with complex $n\times n$ matrices.
Jun
24
reviewed Reject suggested edit on $\pi_1 (x,y) = x$ the projection function with $\pi_1 : R^2 \rightarrow R$
Jun
24
reviewed Approve suggested edit on Expected revenue obtained by the Vickery auction with reserve price $1/2$
Jun
23
revised $SL(V)$ and $PSL(V)$ act $k$-transitively on the space of all $1$-dimensional subspaces.
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Jun
23
answered $SL(V)$ and $PSL(V)$ act $k$-transitively on the space of all $1$-dimensional subspaces.
Jun
23
comment Show $\mathbb{CP^2/CP^1}$ is not a retract of $\mathbb{CP^4/CP^1}$.
$\Bbb CP^2/\Bbb CP^1$ is homeomorphic to the $4$ sphere, and thus its cohomology ring is $\Bbb Z[t]/(t^2)$ with $t$ of degree $4$. If you use the long exact cohomology sequence $$\cdots\to \tilde{H}^*(\Bbb CP^4/\Bbb CP^1)\to H^*(\Bbb CP^4)\to H^*(\Bbb CP^1)\to \cdots$$ you can see that the cohomology ring of $\Bbb CP^4/\Bbb CP^1$ is $\Bbb Z[u,v]/(u^3,v^2,uv=vu=0)$ with $u$ in degree $4$ and $v$ in degree $6$. I'm not certain you can solve the problem only using the ring structure of cohomology.
Jun
23
comment Show $\cos\theta=\frac12(\text{tr}(g)-1)$ with $g\in\text{SO}(3)$
@amWhy upon looking very carefully I did indeed see this ^^ all joking aside, I don't understand the question, but it's fine if all doubts were cleared up by your answer.
Jun
23
comment Show $\cos\theta=\frac12(\text{tr}(g)-1)$ with $g\in\text{SO}(3)$
@amWhy I don't understand what is holding OP back from dividing the equality $\mathrm{tr}(g)-1=2\cos(\theta)$ by $2$...
Jun
23
comment Show $\cos\theta=\frac12(\text{tr}(g)-1)$ with $g\in\text{SO}(3)$
... I don't understand the question.
Jun
23
revised $\exp(\ln(x))=x$ and $\ln(\exp(y))=y$.
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Jun
23
comment What does “coproduct of $\Bbb{Z}*\Bbb{Z}$ of $\Bbb{Z}$ by itself” mean?
The coproduct of $X$ by itself simply means the coproduct of $X$ with $X$. The author is just explaining what they mean by $\Bbb Z*\Bbb Z$.
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.