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Using the Segre embedding $\mathbb P^1 \times \mathbb P^1 \hookrightarrow \mathbb P^3$ given by $(x_0:x_1)x(y_0:y_1) \mapsto (x_0y_0,x_0y_1,x_1y_0,x_1y_1)$, we can rewrite the curve as an intersection of two hypersurfaces in $\mathbb P^3$. Explicitly, these hypersurfaces will be $xw=yz$ and $x^2+y^2+z^2+w^2=xw$. These are both quadrics, so that we have a ...


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You should read $\mathbb Z_p^3$ as $(\mathbb F_p)^3$, that is, the space of vectors consisting of three elements from the field with $p$ elements. This is a vector space over $\mathbb F_p$. (So the elements are not just picked from the numbers $\{0,1,\ldots,p-1\}$; they come equipped with arithmetic operations modulo $p$). There are $p^3-1$ nonzero vectors ...


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Any set of two pairwise disjoint Fano planes is maximal; three or more Fano Planes must share at least one line. Henning's answer is streets ahead in terms of succinctness (and quite clever to boot), but frankly, I didn't do all of this work not to post an answer :) But before I launch into a tedious case-by-case proof that there do not exist $3$ or more ...


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It is not possible to make more than two Fano planes on the same seven points that don't share any line. If one of them is the standard numbering (where three points are collinear iff their bitwise XOR is 0): 4 1 2 7 5 3 6 then the complete set of planes that are compatible with this (up to automorphisms of each plane) is 3 3 ...


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There can be four points on a line. All this says is that there exist four distinct points no three of which are collinear. There might also be other points. EDIT: Let's try thinking of this algorithmically: Let $P$ be the set of points. Step 0. Make a list of all 4-element subsets of $P$. Label them $S_1, S_2,S_3,\ldots,S_n$ (it's a finite list since ...


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Yes, $\alpha$ is invertible; in fact, it's an involution. Let $\lambda_{p}$ and $\rho_{p}$ (for "left" and "right") denote the rulings of the quadric $Q$ passing through a point $p$ of $Q$. The tangent plane $T_{p}Q$ is the unique plane containing $\lambda_{p}$ and $\rho_{p}$. Every "left" ruling (blue) intersects each "right" ruling (green) in exactly one ...


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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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This might be easier to visualize if you identify a point $(x,y)$ in $\mathbb{RP}^2$ with line through $(x,y,1)$ in $\mathbb R^3$ (less the origin). I.e., the projective plane is the plane $z=1$ in $\mathbb R^3$ with points at infinity and the line at infinity added. The $w$-coordinate of the point’s homogeneous representation is just a $z$-coordinate in ...


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An algebraic curve, with affine polynomial equation $p(x,y)=0$, $\deg p=d$, has a corresponding projective curve with polynomial homogeneous equation $P(X,Y,W)=0$, defined as $$P(X,Y,W)=W^d p\Bigl(\frac XW,\frac YW\Bigr)$$ In practice this means that to, say, the affine cubic curve $y^2=x(x^2+1)$, there corresponds the projective cubic curve ...


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Part of the value of thinking about adding points "at infinity" to the Euclidean plane (so constructing the projective plane) and working with the projective plane is that in the projective plane there is symmetry between points and lines. Since each pair of points determines a line, each pair of lines should determine a point. For lines $L$ and $M$ that are ...


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Yes, that's what's meant by $\mathbb Z_p^3$. Each 1-dimensional subspace consists of all scalar multiples of one nonzero point $(a,b,c)$, so it contains exactly $p$ items. Since there are $p^3 -1 $ points to choose, and each subspace contains $p$ points, you might think you'd get $(p^3 -1 )/p$ subspaces. . That's wrong because $(0,0,0)$ is in every ...


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EXAMPLE: Consider the Fano Plane: Condition (iii) of the definition says: there are four points no three of which lie on a line. Now, points $1,2,3$ do lie on a line, however, consider the points $2,3,6,7$. At most two of these points lie on a line. No three of the points $2,3,6,7$ lie on a line. Thus, condition (iii) is satisfied. Another example ...


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Projective equivalence means that there exists a projective transformation transforming one into the other. A projective transformation in $d$-dimensional projective space is uniquely defined by $d+2$ points and their images, both in general position. So you can choose any polytope with more than $d+2$ vertices, and if you move one vertex a bit then they ...


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$\newcommand{\dd}{\partial}\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$I don't have access to the book, but the indexing appears to be off. (There should be $X_{j}$ instead of $X_{i}$ on the right side of $v_{i}$.) Assuming that's correct: If $A = [\alpha_{ij}]$ is an invertible $(n + 1) \times (n + 1)$ complex matrix, $X = (X_{0}, \dots, ...


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I'm confused as to how you want the duality here. Let $V$ be the underlying vector space, let $k$ ($=\mathbb{R}$ or $\mathbb{C}$) be the field over which the vector space is taken (so $V=k^3$). Let $(x_1:x_2:x_3)$ be homogeneous coordinates in the projective plane $\mathbb{P}^2$ (defined by the standard basis of $V$). Let $p_1=(\alpha_1:\alpha_2:\alpha_2)$ ...


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You can visualize this action very explicitly: a tangent vector to a point $p \in S^2$ is literally a little vector tangent to $S^2$ inside of $\mathbb{R}^3$, and rotation acts in the obvious way. The rotations around the axis through $p$ act transitively on unit tangent vectors at $p$ (and so act transitively on the projectivized tangent space at $p$), and ...


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Hint: Assume we have Fano${}_1$ and Fano${}_2$ 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', starting out from these three points.



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