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4

We have by a simple induction $T^n=T$ so $$\exp (T)=\sum_{n=0}^\infty \frac{T^n}{n!}=I+T\sum_{n=1}^\infty\frac1{n!}=I+(e-1)T$$


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In this post I worked out, as an exercise, the answer for smooth projective complex hypersurfaces, not using de Rham cohomology. After applying the Lefschetz hyperplane theorem, the basic tool is characteristic classes.


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Every Möbius transformation is an automorphism of the extended plane. So $Tz\in \mathbb{R}\cup\{\infty\}$ if and only if $z\in T^{-1}(\mathbb{R}\cup\{\infty\})$. Since Möbius transformations map circles (in the extended plane) to circles, that means $z$ lies on a circle passing through $z_1,z_2,z_3$. But through any three points in the extended plane, there ...


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Without additional reference objects, this is impossible: any pair of a point and a line can be mapped to any other such pair by a projective transformation, so any projectively invariant quantity you assign to that has to be constant, i.e. be the same for all pairs, in which case I wouldn't call it a distance. If you introduce additional reference objects, ...


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I will sketch the argument, since writing all the details would be too tedious, and would certainly include typos... An important idea about blowups of points in the plane is the following: given two curves passing through a point $P$, the blowup $\operatorname{Bl}_P(\mathbb P^2)$ separates the curves (i.e. their strict transforms do not intersect) iff they ...


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The geodesics on $H^2$ are, indeed, semi-circles centered on the real line and also vertical half-lines. This is the hyperbolic (curvature=-1) plane, so the only projective axiom which can be violated is the second one - and, indeed, there are lines which do not intersect (e.g., semi-circles with the same center and different radii).


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It appears my comment was enough to answer the question, so I'll promote it to an answer: Projective space doesn't have an origin. To elaborate a little - the origin of a vector space is special because it is the additive identity under the group structure. But once you projectivise the vector space, the group structure disappears - and there is no ...


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Use the series representation of the exponential $$ e^T = \sum_{n=0}^\infty \frac1{n!} T^n, $$ from $T=T^2$ you can deduce $T^n=T$ for all $n$.


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Edit: This is just a worked out example of a blow-up. A solution was posted by Andrew here. I will work out the blow up for the function $f(x,y) = y^2 - x^3 - x^2$. Note that $\nabla f(0,0) = 0$, so there is a singularity at $(0,0)$. If you graph the function (over $\mathbb{R}$), you will see that it crosses over itself at $(0,0)$. Consider $f(x,xy) = ...


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The algebraic projective variety $V\subset \mathbb{P}^n$ is given by the zero locus of homogeneous polynomials $f_i\in \mathbb{C}[x_0,\dots,x_n]$. Take now the open subset $U$ of $\mathbb{P}^n$ where $x_0\not=0$, which is an affine space. You can assume that $x_0=1$ and obtain then coordinates $x_1,\dots,x_n$. This gives an isomorphism ...


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There is a general algorithm for computation of (co)homology groups of real-algebraic subsets in $\mathbf{R}^n$. Being a hypersurface does not particularly help in this computation. The algorithm goes back to Tarski's work (on elimination of quantifiers). The entire book Algorithms in Real Algebraic Geometry is pretty much all about such computations. ...


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HINT: Consider the plane spanned by $\mathfrak L$ and $\mathfrak P$. A lovely elementary book is Pedoe's Geometry: A Comprehensive Course.


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A point $p$ is incident with a set $S$ iff $p\in S$. Given how we define the image of $S$ under a map, how could a map fail to preserve incidence?


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The way I read it, the book does not say that $AB\cap A'B'=AD\cap A'D'$. What it does say is this: But these points are projected for $X$ into $P,Q,R$, on $\Sigma$, hence $P,Q,R$ are collinear. Do you have some point $P'=AD\cap A'D'$ and $X,P,P'$ are collinear, so if you project $P'$ into $\Sigma$ with $X$ as the center of projection, then you end up ...



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