# Tag Info

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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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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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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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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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$V$ 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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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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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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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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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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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: $$A= \begin{bmatrix} 1&0&0\\ 0&1&0\\ ... 1 Each line has just one point at infinity, which is approached by going in either direction along the line. Two lines share the same point at infinity if and only if they are parallel to each other. Two lines not parallel to each other have different points at infinity. When one adds to the affine line a point at infinity that is approached by going in ... 1 I'm sure I've seen this phenomenon before: one starts with some quantity defined in terms of coordinates and asks how anyone would think of that and how it could be natural. This is too "up close." For example, is it better to understand the determinant of a transformation on a real vector space as a change in volume, or is it more valuable to dig into a ... 1 You want to use the fact that x^{31} = 1 for all x \neq 0, so you compute the exponents \bmod{31}. So, for example, 5^{-1} = -6 = 25, and 6^{-1} = -5 = 26. 1 Let \ell be the line containing A,B,C and m the line containing a,b,c. Take any line n different from \ell and m. Then the intersection points Aa\cap n, Bb\cap n, and Cc\cap n are collinear. 1 Essentially, no. Pappus and Desargues, at a deep level, follow from the properties of linear systems of cubic curves in the projective plane. By a counting argument, any cubic that contains eight of the nine intersection points of two fixed cubics must necessarily also contain the ninth intersection point as well. This and the fact that a degenerate cubic ... 1 The matrix for the first conic is$$ \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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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 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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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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