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1

More generally, any reducible variety is either disconnected or singular (or both!). This is because if there are two components that intersect nontrivially, a point in the intersection would not have its local ring be a domain, much less a regular local ring. Now you have a reducible cubic surface which is necessarily not disconnected because two ...


2

Suppose for contradiction that some line $l$ lies on $Q$ ($l\subset Q$) but does not go through the apex (=singular point) $S$ of $Q$. Then the whole plane $\Pi$ determined by $S$ and $l$ is contained in $Q$ ($\Pi\subset Q$): indeed for any $L\in l\subset Q$ the line $\overline {SL}$ lies on $Q$ and $\Pi$ is the union of all those lines $\overline {SL}$ ...


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 ...


0

Angular speed is calculated by, $$v=\frac{\Delta{\theta}}{\Delta{t}}$$ We know that, $$\theta=\frac{s}{r}$$ Here, $r=f$ $s=m\widehat{AB}$ $\theta$ in radians. Since you are not given any numerical value, your answer will only have constants (alphabetical) and variables.


0

The calculation isn't much different from just moving in a straight line. The only difference is how you calculate the distance travelled. Provided you know the change in angle $\theta$, then the distance travelled is $d=\theta\cdot f$ so $$v=\frac{\theta\cdot f}{t}$$


0

Map to the unit disk so that $z_1$ goes to center. Rotate the disk so that the image of $z_2$ is real. Map back to the upper half plane with $z\mapsto i\dfrac{1+z}{1-z}$. The composition of the above maps achieves the goal. Any Möbius transformation of the upper halfplane onto itself is represented by a $PSL(2,\mathbb{R})$ matrix; Wikipedia.


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 ...


0

Here I am going to prove the first formulation of the statement in completely non-analytic way. In this proof, by "sides" of a triangle I will mean lines containing these sides. I will first prove a lemma: let $A_1A_2A_3$ is any triangle and let $B_1,B_2,B_3$ be points on sides of this triangle opposite to $A_1,A_2,A_3$ respectively, such that ...


0

Any two ideal triangles are congruent, or in other words, any non-degenerate triangle inscribed into the unit circle can be mapped to any other by a projective transformation which fixes the unit circle. So without loss of generality you can restrict your considerations to one special case. I like coordinates, so I'd start with $\triangle A_1A_2A_3$, the ...


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 ...


0

Fortunately, both squares are in the same plane. At first, I thought perhaps they were on separate planes. Let the square at bottom left be denoted $ABCD$, with $A$ at lower left and running counter-clockwise around from there. Let the other square be denoted by $CEFG$, with the letters again running counter-clockwise. The lower diagram is a top-down, ...


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 = ...


0

One possible answer: A projective transformation preserves collinearity of points. A degenerate conic contains collinear points, a non-degenerate does not. So these two cannot be projectively equivalent. A second answer: if you imagine the embedding in 3-space, then your projective plane can be visualized as the intersection of an affine (drawing) plane ...


0

Hint By construction, the standard action of the special orthogonal Lie group $SO(Q)$ on $V$ induces a transitive action on $\Sigma$.


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} & ...


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 ...


0

In general you have three degrees of freedom when mapping one conic to another. One approach is the following: choose three distinct points $A,B,C$ resp. $A',B',C'$ on each conic. Then construct a point $D$ as follows: take the tangents to the first conic in $A$ and $B$, intersecting these in a point $P$ and join that point of intersection with $C$. That ...


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

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 ...


0

Yes, this is true, and follows form Bezout's theorem. Without loss of generality, we can assume that $\infty = (0:0:1)$ is on the curve (call this the point at infinity). Let the singular point be called $S$. Now choose a line $L$ in $\mathbb P^2$ that does not pass through $S$. Define a ma from this line to the curve as follows: for each point $P \not \in ...


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, ...


1

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



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