# Tag Info

5

As I start writing, I'm not convinced it's even a $4$-polytope. It probably is, but I'll at least attempt a decomposition into $3$-faces and hopefully compute the Euler characteristic. I don't believe it's a well-known polytope, and I think the 'outer' facet is actually a truncated octahedron with hexagonal pyramids attached to each square face (there are ...

4

Go up one dimension and into affine space. There $P$ corresponds to the line $L = (tb_0, tb_1,\ldots,tb_n)$ and $H$ still has the same equation. Now apply a linear transformation to $\Bbb A^{n+1}$ that moves $H$ to be orthogonal to $(1, 0,0,\ldots,0)$ (remember that $H$ originally is orthogonal to $(a_0, \ldots,a_n)$) (this might move $L$ to someplace ...

4

Observe, in the Segre embedding $\mathbb{P}\times\mathbb{P}\rightarrow\mathbb{P}^3$, we have $$(x_0:x_1)\times(y_0:y_1)\mapsto(x_0y_0,x_0y_1,x_1y_0,x_1y_1).$$ If we choose the coordinates on $\mathbb{P}^3$ to be $(x,z,w,y)$, we see that $xy=zw$ on this surface.

2

See this blog post. It's true in general that the cohomology of a smooth hypersurface of degree $n$ in $\mathbb{CP}^d$ only depends on $d$ and $n$.

2

"Projection" can mean many things, and about the only feature they have in common is that it's about an (idempotent) transformation that moves points somewhere along straight lines. (And even that is not fully general -- although some "map projections" can be explained as moving points along straight lines from the surface of a sphere to an image plane, ...

1

The answer depends on what you mean by "bad in geometry." For many people the problem is very simple. Geometry isn't taught in schools the way it once was, so you can be very good at algebra and other things, but geometry can still seem foreign and strange. It might be good to learn geometry - at first - in normal school books from earlier times in order ...

1

There is a projective transformation that takes any line $L$ to $\mathbb V(x)$. And there is a projective transformation that takes any nondegenerate conic $C$ to $\mathbb V(x^2 + y^2 + z^2)$. The first is just linear algebra, and the second can be proved using your result before and the fact that there is a unique conic passing through any $5$ general ...

1

Here's a geometric proof. Let $O$ denote the origin. If $P$ and $Q$ are two points on the unit sphere, then $\sin(\angle POQ)$ is twice the area of triangle $\triangle POQ$. Thus, it suffices to prove that $$\text{area}(\triangle POQ) \,+\, \text{area}(\triangle QOR) \;\geq\; \text{area}(\triangle POR)$$ for any three points $P$, $Q$, and $R$ on the ...

1

A probably-irrelevant remark: if you define the distance to be the sine of half the angle between the points (required to be in the same hemisphere!), then the triangle inequality is straightforward, because $2 \sin \frac{\theta}{2}$ is exactly the length of the chord from $P$ to $Q$, so the thing you're trying to prove becomes...the Euclidean triangle ...

1

It depends on how exactly you define perspective projection, i.e. which properties are part of the definition and which are consequences thereof. But you can think about this geometrically. You have a line in space, and an eye point. Assuming that the line didn't pass through the eye, connecting the eye to every point on the line yields a bundle of lines, ...

1

The whole point of the representation you're using for affine transformations is that you're viewing it as a subset of projective space. A line has been chosen at infinity, and the affine transformations are those projective transformations fixing this line. Therefore, abstractly, the use of the extra parameters is to describe where the line at infinity ...

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