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8

Both of these implications are incorrect. As $\mathbb{CP}^n$ is Kähler and complex submanifolds of Kähler manifolds are Kähler, all projective manifolds are Kähler. The converse however is not true. That is, there exist Kähler manifolds which are not projective. For example, all two (complex) dimensional tori $\mathbb{C}^2/\Lambda$ are Kähler, but many of ...


3

This is because you are free to fix the coordinate system that suits you most. Let $V_1$ and $V_2$ the linear subspace in $\Bbb A^{2n+2}$ inverse images of $H_1$ and $H_2$ respectively under the canonical quotient map $$ \Bbb A^{2n+2}\setminus\{0\}\longrightarrow\Bbb P^{2n+1}. $$ Then $\dim(V_1)=\dim(V_2)=n+1$ and choose coordinates with respect to a basis ...


2

If I understood your question well, it is enough to exhibit $k$ independent points on the hyperplane at infinity: then their span is the hyperplane at infinity and you just need to intersect this hyperplane with every line joining the origin and one of the unit versors. To do it, consider the plane spanned by the origin and $k-1$ of the unit versors, say ...


2

I've only got time to outline this, but here we go... We have subspaces $U, V, W$ of $K^{n+1}$, of dimension $\tilde k = k+1, \tilde l = l+1$, and $\tilde m = m+1$ respectively, with $\tilde k + \tilde l + \tilde m \geq n + 2$. First, note that if any two subspaces have a non-zero vector in common, we are done, because their projectivisations intersect, ...


2

If you call the homogeneous coordinates of ${\mathbb P}^2$ by $(X:Y:W)$ you must homogenize $x^2-y$ to $X^2 - Y W$ (set $x=X/W$ and $y=Y/W$ and take the numerator of the resulting rational expression). Now substitute $Y= Y_1-W_1$ and $W=Y_1+W_1$ and $X=X_1$). Then $X^2-Y W$ goes into $X_1^2 - (Y_1^2 - W_1^2) = X_1^2 + W_1^2 - Y_1^2$. This is $x_1^2 + w_1^2 = ...


2

Have a look at this paper by Sirvent and Wang. It's not elementary but it defines the process quite completely with a worked example. In particular, the contracting space ${\mathbb H}_c$ is clearly defined. On slide 3 of the presentation, it simply says that ${\mathbb H}_c$ is "generated by the eigenvectors of $\beta$ Galois conjugates". (Of course, it's ...


1

One possible solution to the problem is to use the Schubert calculus in the grassmannian $\mathbb G=\mathbb G(1,\mathbb P^n)$ of lines in $\mathbb P^n$. That grassmannian has dimension $2n-2$ and the set of lines intersecting $\mathbb P(U)$ is a subvariety $\Omega _{n-1-k}$ (this is a standard notation) of codimension $ n-1-m$ in $\mathbb G$. The lines ...


1

Indeed there is a theorem which say: Let $V,W$ be $(n+1)$-dimensional vector spaces further on let $P(V),P(W)$ be two $n$-dimensional projective spaces and assume that $$ A_1, \dots,A_{n+2} \in P(V) \quad \text{and} \quad B_1, \dots,B_{n+2} \in P(W) $$ are in general position. Then there exists a unique projective transformation $f \colon P(V) \to ...


1

Adding to the previous answer one might notice that all complements of hypersurfaces $\mathbb{P}^n_k - V(f)$ where $f$ is homogeneous in $S=k[x_0,\ldots,x_n]$ of degree $d > 0$ are affine open subsets. Scheme theoretic this is just the isomorphism $\mathrm{spec}(S_{(f)}) = D_+(f)$ where $D_+(f) = \{\mathfrak{p} \in \mathrm{proj}(S) \mid f \notin ...


1

$\newcommand{\Proj}{\mathbf{P}}$An "affine open subset" of $\Proj^{n}$ normally refers to one of the $(n+1)$ sets you mention. That said, the projectivization $\Proj(V)$ of a vector space $V$ doesn't come with coordinates (in the same way $V$ itself doesn't come with a preferred basis), so an "affine open set" could conceivably refer to an arbitrary ...


1

You are correct that $[0:1:0]$ is not on the curve $y^2=x^3+Axz^4+Bz^6$ in those modified projective coordinates $[x:y:z]$, and $[1:1:0]$ is indeed on the curve. I am not an expert though in cryptography... the literature in cryptography seems to insist in calling the point at infinity by $[0:1:0]$ even though this point is not on the curve in those ...


1

The morphism $\varphi_C$ is best seen as a morphism $\varphi_C:C\rightarrow(\mathbb{P}^2)^*$ from the curve $C\subset \mathbb P^2$ to the dual projective space $(\mathbb{P}^2)^*$, whose typical point $Q=(a:b:c)\in(\mathbb{P}^2)^*$ corresponds to the line $l_Q\subset \mathbb P^2$ of equation $ax+by+cz=0$. If $C$ is a line $ax+by+cz=0$, then the morphism ...


1

If you try the $n=2$ case first, you'll quickly see that all the others are nearly the same. When $n=2$ we have a four-dimensional vector space, say $$V = \operatorname{span}\{e_0, e_1, e_2, e_3\}.$$ Since every nondegenerate quadratic form on a vector space (in characteristic $\neq 2$) is equivalent to every other, we can just pick a quadratic form and ...


1

The metric you described is the standard metric on the projective space: in the real case it can be visualized as the angle between lines (thinking of the elements as lines). It arises as the quotient of the spherical metric on $S^n$ by the group of isometries $\{x\mapsto \alpha x, \ |\alpha|=1\}$ where $\alpha$ belongs to the ground field, $\mathbb{R}$ or ...


1

Consider the direct product of two projective lines or, better, of a projective line and projective n-space for any n.


1

A conic is described by a symmetric $3\times3$ matrix, and multiples of a matrix describe the same conic. The polar line of a point can be obtained by multiplying that matrix with the point, but the result will only be unique up to scalar multiples. So you essentially have two vector equations: \begin{align*} \lambda a &= \begin{pmatrix} m_{11} & ...



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