New answers tagged

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Hint: Assume we have Fano1 and Fano2 given, and pick and fix any three noncollinear points in both, call them e.g. $O_1,X_1,Y_1$ and $O_2,X_2,Y_2$. Try to express the lines and the further points in terms of 'intersections' and 'lines on two points'.


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To each lie algebra $\mathfrak{g}$ there is a unique simply connected Lie group $G$ having $\mathfrak{g}$ as its lie algebra, and furthermore any other Lie group $H$ having lie algebra $\mathfrak{g}$ is covered by the universal one $G$, in other words is a quotinet of $G$ by some discrete central subgroup $K$. (In fact, covering space theory goes on further ...


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Here's the solution I have eventually employed: Assume that there are spaces A and B, which are related by homography (mapping stuff from A to B): $$ H = \begin{bmatrix} h_{11} & h_{12} & h_{13}\\ h_{21} & h_{22} & h_{23}\\ h_{31} & h_{32} & h_{33}\\ \end{bmatrix}\quad. $$ As $H$ is general homography, some of the points in A may ...


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This follows easily from representation theory of group $G = Spin(10)$ (the simply connected group of type $D_5$). The spinor variety is the homogeneous space $G/P$, where $P$ is the maximal parabolic group corresponding to the simple root $i_5$ (for the standard enumeration of vertices). The 16-dimensional spinor representation --- the space of global ...


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You can consult the following article: W. Lichtenstein "A system of quadrics describing the orbit of the highest weight vector", Proc. AMS 84 (1982), no. 4, 605–608, where it is proved that homogeneous varieties in their minimal embedding have ideal generated by quadratic forms. You can deduce that they are ten going to lines section and finding a canonical ...


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I get $G=\mathrm{GL}(n,\mathbb{C})$, let $B$ be a Borel subgroup of $G$, let $P$ be a parabolic subgroup of $G$ and let $T$ be a maximal torus in $G$; we know that: up to conjugation: $T$ is the group of diagonal matrices, $B$ is the group of upper triangular matrices; in particular: $B$ is $T\ltimes U$, the semidirect product of $T$ with the group of ...


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This answer of mine shows the general approach of how to compute a projective transformation in the plane given four points and their images. Just plug your numbers into that algorithm and you end up with the required transformation matrix. How to show that a projective transformation is unique depends on what definitions you have established. Usually you ...


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If you have established that cross ratios are invariant under projective transformations, you can choose three of the points without loss of generality. E.g. make three of them $0,1,\infty$. Then you'll only have to deal with a single real parameter. If you haven't established this invariance yet, you may start by doing so. Another attempt would be writing ...


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The matrix for the first conic is $$ \begin{bmatrix} 1 & -1 & 2 \\ -1 & -8 & 1 \\ 2 & 1 & 4 \end{bmatrix} $$ which has determinant $-9$. The matrix for the second conic is $$ \begin{bmatrix} 1 & -2 & -1 \\ -2 & 1 & 0 \\ -1 & 0 & 0 \end{bmatrix} $$ which has determinant $-1$. So neither is singular.


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You want to use the fact that $x^{31} = 1$ for all $x \neq 0$, so you compute the exponents $\bmod{31}$. So, for example, $5^{-1} = -6 = 25$, and $6^{-1} = -5 = 26$.


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The Grassmannian is the collection of all subspaces of a projective space having a fixed dimension. It can be represented using Plücker coordinates as an algebraic variety, which essentially lets you embed the Grassmannian into a projective space. For example, if you have $V$ being a 4-dimensional vector space, then $P(V)$ is a 3-dimensional projective ...


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I'm sure I've seen this phenomenon before: one starts with some quantity defined in terms of coordinates and asks how anyone would think of that and how it could be natural. This is too "up close." For example, is it better to understand the determinant of a transformation on a real vector space as a change in volume, or is it more valuable to dig into a ...


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It depends on the background. If you are defining Euclidean geometry based on the imaginary circle points $(1,\pm i,0)$ (along the lines of a Cayley-Klein metric), then it is possible to replace these two complex points by two real points, e.g. $(1,\pm 1,0)$, and study the resulting pseudo-Euclidean geometry. I know Perspectives on Projective Geometry has ...


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projective transformations Any matrix, usually excluding singular matrices (determinant zero). That's the most general case. affine transformations Preserve points at infinity, i.e. result has $z=0$ if and only if input has $z=0$. That is what you describe with $h_{31}=h_{32}=0$ and $h_{33}=1$, except that last coordinate $h_{33}$ can in fact be ...


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It's mostly a matter of convention. In my surroundings, one would define $$CR(A,B;C,D):=\frac{[A,C][B,D]}{[A,D][B,C]}$$ which is the same order Wikipedia uses. I assume that convention to be the most common one. But I, too, have seen other conventions, so it's best to check what convention a given source is using. I'd use $[A,C]$ to denote a determinant ...


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$V$ is certainly changing in the sense that after applying a coordinate transformation $A$ to $\mathbb{P}^n$ you will find that $A(V)$ is not (in general) the zero locus of the same homogeneous polynomials $F_1,\ldots,F_m$ that $V$ was, but the author is saying that there still exist other homogeneous polynomials $G_1,\ldots,G_m$ that cut out $A(V)$ as their ...


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The statement only holds for the one-dimensional case, the projective line. One way to prove it is by showing that the cross ratio remains invariant under projectivities, and verifying that given three points, the cross ratio already uniquely defines a fourth. Another approach would be by observing that a projectivity in $d$ dimensions is uniquely defined ...


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The situation is much simpler to conceptualize if we align our coordinate system so that the light rays are parallel to a coordinate axis: Left: $\varphi \lt \beta$. Right: $\varphi \gt \beta$. Above, $S$ is the distance between the pivot points, $L$ is the width of the louvres, $\beta$ is the solar elevation angle (the angle between horizontal and the ...


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There’s a model of the projective plane that might be helpful to you. Consider the plane $z=1$ in $\mathbb R^3$. A point on this plane with coordinates $(x,y,1)$ can be identified with the line through the origin that intersects the plane at this point, so that any non-zero multiple of $(x,y,1)$ represents the same point. If you then add the lines parallel ...


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Just speaking in "intuitive" terms, consider three lines intersecting at one point. Keep one line firm, tilt one of the line, making the intersection to move away towards infinity, in one of the directions, slide the third line parallel to itself to keep the one-point intersection. Continuing to tilt beyond the parallelism of the first two lines, the one ...


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Each line has just one point at infinity, which is approached by going in either direction along the line. Two lines share the same point at infinity if and only if they are parallel to each other. Two lines not parallel to each other have different points at infinity. When one adds to the affine line a point at infinity that is approached by going in ...


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Multiply by $y(x^2+1)$ so your map is $((x^4+3y)y, (x+1)(x^2+1), y(x^2+1))$. Now further you want $X,Y,Z$ to be homogeneous so set $x=\frac{X}{Z}$ and $y=\frac{Y}{Z}$ and now multiply by the lowest power of $Z$ to clear the denominators.


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One considers all non-zero multiples of $(x,y,1)$ to form an equivalence class. If you were to include the null vector in that class, then you'd have to include it in the class for every point. This breaks transitivity. In other words, without the null vector you can say that if $a\sim b$ (they are the same point), and $b\sim c$, then $a\sim c$. If $b$ could ...


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If you can't see the answer to the first half of your question, then you really have not understood the terms being used. Consider: The points of $P(V)$ are the lines of $V$. You make a linear transformation of $V$ into a projective transformation$T$ on $P(V)$ by mapping the projective point $[v]$ to the projective point $[T(v)]$. An eigenvector is just ...


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Essentially, no. Pappus and Desargues, at a deep level, follow from the properties of linear systems of cubic curves in the projective plane. By a counting argument, any cubic that contains eight of the nine intersection points of two fixed cubics must necessarily also contain the ninth intersection point as well. This and the fact that a degenerate cubic ...


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You're right to worry about this; in more general settings ($k$ not a field but a more general commutative ring) the natural map $\text{Aut}(V) \to \text{Aut}(\mathbb{P}(V))$ might fail to be surjective, so it won't be possible to use this definition. Unfortunately, the correct definition at this level of generality is relatively sophisticated (even ...


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I think the simplest counterexample is if you take $C$ to be the standard twisted cubic, viz. the image of the closed embedding $f\colon\mathbb{P}^1\to\mathbb{P}^3$ given by $[U:V] \mapsto [U^3:U^2V:UV^2:V^3]$ (if we call $[T:X:Y:Z]$ the coordinates on $\mathbb{P}^3$, it has equations $TY=X^2$, $XZ=Y^2$, $TZ=XY$, but no matter). This morphism is an ...


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Let $\ell$ be the line containing $A,B,C$ and $m$ the line containing $a,b,c$. Take any line $n$ different from $\ell$ and $m$. Then the intersection points $Aa\cap n$, $Bb\cap n$, and $Cc\cap n$ are collinear.


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Let $L_i$ be the lines. Take $P_1\in L_1$ and consider the unique plane $\Pi$ passing through $P_1$ and containing $L_2$. That plane does not contain $L_3$ : else it would contain both $L_2$ and $L_3$ and since two lines in a projective plane always intersect, $L_2$ would intersect $L_3$, contrary to the skewness assertion. So $\Pi$ cuts $L_3$ in just one ...


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@MvG has answered it and I accepted it. I just wanted to post a picture I've drawn for this to help me understand in case if someone else finds it useful.


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Consider three-dimensional space $\mathbb R^3$. If you normally homogenize by appending $z=1$, that means that your geometry as you know it happens on the $z=1$ plane in space. But it's also possible to view geometric elements as linear (i.e. containing the origin) subspaces of the whole three-dimensional space. A point on the plane corresponds to a line ...


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Given a homogeneous polynomial $f(x,y,z)\in \mathbb C[x,y,z]$ of degree $d\gt0$, the curve $V(f)\subset \mathbb P^2$ (= the zero-locus $\{f=0\}$) has as singularities the set of solutions $S \subset \mathbb P^2$ of the system $$\frac {\partial f}{\partial x}=\frac {\partial f}{\partial y}=\frac {\partial f}{\partial z}=0 \quad (\ast)$$. Amazingly a ...


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Recall that a plane in $\mathbb R^3$ is a line in $\mathbb P^2(\mathbb R)$ and that a line in $\mathbb R^3$ is a point in $\mathbb P^2(\mathbb R)$. The plane $x_0 = 0$ in $\mathbb R^2$ corresponds to the projective line parametrized by $(0:s:t)$ for $s,t \in \mathbb R$. Similarly the plane $x_0 + x_1 + x_2 = 0$ in $\mathbb R^2$ corresponds to the ...


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Let $ \pi : V \backslash 0 \to \mathbb P(V)$, $v \mapsto [v]$ the projection. If $U,U'$ are two subspaces, the subspace $U+U'$ is precisely generated by the sums $u + u'$ where $u \in U, u' \in U'$. Take any line (through the origin) $\ell$ contained in $U+U'$. $\ell$ can be written as $t\xi$ for some vector $\xi \in U + U'$. We want to show that there ...


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I don't think this is true. Consider the example where $X$ is cut out by $X^2+Y^2+Z^2=0$ in $\mathbb{P}^2$ and the map $X\rightarrow \mathbb{P}^1$ given by $[X:Y:Z]\rightarrow [X:Y]$. If you fix $[X:Y]\in \mathbb{P}^1$, the preimage will consist of two points (unless $[X:Y]=[1:I]$ or $[1:-I]$), so the map is degree 2. In general, I think the degree $d$ ...


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If I well understand your question, the answer can be done using homogeneous coordinates. Given a point $P=(a,b)$, his homogeneous coordinates are $P=[a,b,1]^T\equiv [ca,cb,c]^T$ ( see here for a definition). using this the projection from the origin on the line $x=1$ can be represented by the matrix: $$A= \begin{bmatrix} 1&0&0\\ 0&1&0\\ ...


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By applying a projective transform that moves $P$ to infinity and preserves symmetry about the line $\ell$, it is sufficient to prove the statement in the limit case when $P$ is the ideal point of $\ell$. In the limit case the lines $A'A_1$, $B'B_1$ and $C'C_1$ are parallel to $\ell$. For every point $X$, let $d(X)$ denote the signed distance between $\ell$ ...



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