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olivier dot begassat dot cours at gmail dot com


Jun
29
comment Linear Algebra without Matrices
Matrices allow you to make calculations mechanical, and to state results concisely. There are definite instances where matrices save you times and ink.
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
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
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
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
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.
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
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?
$\exp{}{}{}{}{}$