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Many of these are proved using the universal hypersurface. Let $P$ the projective space of all degree $d$ forms (since the equation and any non-zero constant multiple give the same variety) and consider $Z\subset P\times\mathbb{P}^n$ the universal hypersurface defined in the obvious way - these are pairs $(f,p)$ with $f(p)=0$. Consider $T\subset Z$, defined ...

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The equations which define $P_3$ have coefficients in $k$, so $Gal(\bar k/k)$ preserves the locus of these equations i.e it fixes the coordinates of $P_3$ since it is the unique point of this locus.

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I think it’ll help if we restrict our view to the finite plane, i.e. setting $Z=1$, using the equivalence $(x,y,z)\leftrightarrow(x/z,y/z,1)$, which is good except on the “line at infinity” given by $Z=0$. You’ve done that in your exposition. Now, to modify slightly what you’ve said, the given map $\phi$ has the effect of sending $(x,y,1)$ to ...

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$f$ is homogeneous and the denominator is never zero, so it gives a well-defined function on $\mathbb{P}^n$. It is clear that a matrix $A$ in the image of $f$ has trace $1$, is symmetric, and the fact that $A^2 = A$ is an easy algebraic verification. In the coordinate patches given here, away from, for example, $x_{n+1} = 0$, the map is given locally as ...

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Suggestion: Get a copy of Hartshorne's little book on Projective Geometry. It'll answer questions like this. Your question seems to me "How do we know that any two lines have the same number of points?" (since whether we call this "$n$" or "$n+1$" really doesn't matter). Here's a quick proof, starting with Hartshorne's four axioms for projective planes. ...

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I will use the following notation The definition of the Veronese map will be taken from the question above. I will denote the coordinates in $\mathbb{P}^{5}$ by $[z_{0}:\ldots:z_{5}]$ and the coordinates in $\mathbb{P}^{2}$ by $[x_{0}:x_{1}: x_{2}]$. Claim The Veronese surface is cut out by three quadrics: $$C_{1}: z_{0}z_{3} - z_{1}^{2} = 0$$ $$C_{2}: ... 1 These three lines form the parallel class of lines having slope 2. I think they are shown as kind of weird in this drawing, but it is arbitrary what order they are in. For example, consider y = 2x, this line is satisfied by (0,0), (1,2), (2,1). These three points make up the blue line in the picture. None of these three lines intersect, and they ... 1 This is a proof by homogeneous coordinates, projective geometry and computer algebra. I start with the first case. First off, observe that your whole setup is invariant under projective transformations. So without loss of generality you may assume that the original conic is the unit circle, and you may further assume that P is at position [1:0:1]. Then ... 1 This is a svd decomposition, and it's not unique, because two singular values are equal. That means that U and V are not unique, and probably not in the most symmetric form. The same degeneracy may be present in the solution itself, although the absolute value in the condition on component may then push this into an awkward angle. The fact that the same ... 1 You can tweak U and V so that A becomes \left[\begin{array}{ccc}a&a&a\\a&a&b\\a&b&c\end{array}\right]. (a=0.9595767,b=-0.4841173,c=0.0808466) Then A^TA=A^2 has eigenvalues 1,1,4, so A has eigenvalues \pm1,\pm1,\pm2 The choice 1,-1,2 gives 2a+c=2,a(a-b)^2=2 and I think: 2ac-b^2-a^2=-1. I think it turns into a ... 1 First of all, the map is clearly not a map from \mathbb C^3 to \mathbb C^2, but a map from C^3\setminus\{(0,t_2,t_3)|t_2,t_3\in\mathbb C\} to \mathbb C^2. The map is clearly undefined on (0,1,1) for example. Now, if \phi(t_1,t_2,t_3) is defined, then you can find the kernel using this hint:$$\frac{t_2}{t_1} = 0 \iff t_2 = 0$$1 The smooth and topological answers should agree so I will just concentrate on the topological case. Let me restrict attention to the complex case although the real case looks similar. The classifying space of \mathbb{P}^{n-1} bundles is BPGL_n(\mathbb{C}). There is a natural functor from rank n complex vector bundles to \mathbb{P}^{n-1}-bundles and ... 1 Let ABCD be your square. Let's call G and H the points where l intersects, respectively, the sides AB and CD. We have the two trapezia GBCH and AGHD that both have the same area (half of the original square, so 1/2). We now want to reflect AGDH around l, let's call L and K the images of A and B after the reflection. The new ... 1 HINT You want to reflect lines  x= \pm 0.5, y= \pm 0,5  about  y = m x , m= \tan \theta . Instead of that consider four among the eight rotated lines$$ \pm x \cos \theta \pm y \sin \theta = 0.5 ; \, \pm x \sin \theta \pm y \cos\theta = 0.5 $$1 What definition of irreducible are you using? The one I have in mind, which is also the definition given in Hartshorne, is: Definition: A nonempty subset of a topological space Y \subseteq X is irreducible if it cannot be expressed as a union Y = Y_1 \cup Y_2 of two proper subsets each of which is closed in Y. Thus, we say that a reducible space ... 1 The Cartesian equation of the circle of radius r and center (a, b) in the (x, y)-plane is$$ (x - a)^{2} + (y - b)^{2} = r^{2}. \tag{1} $$To "homogenize" in projective coordinates (w, x, y), multiply out, then multiply each monomial by the appropriate power of w to make each term have the same total degree in (w, x, y). Here, (1) becomes$$ ...

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