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6

If you're using the usual topological description, then the short answer is yes: We can describe the projective plane as the quotient of $S^2$ (the sphere) by the action of $\{\pm 1\} \cong \mathbb{Z}/2\mathbb{Z}$. As the quotient of a compact space it is automatically compact (continuous image of a compact set). As a comment, there isn't only one way to ...

2

I have found my answer to this question in this paper of F. L. Zak which contains the bound $$b(X) < {{N-1}\choose{\lfloor\frac{N-1}{2}\rfloor}} d^N$$ for a nonsingular variety $X \subseteq \mathbf P^N$ defined by equations of degree $\leq d$.

2

You are basically there. You just need to determine $\lambda$. Given an arbitrary $[z_0,\dotsc,z_n] \in \mathbb{CP}^n$, you want to find $\lambda$ such that $$\sum_{k=0}^n \lvert \lambda z_k\rvert^2 = 1,$$ and $\lambda\cdot z_n \geqslant 0$. These conditions are easy to satisfy.

2

The answer to your first question is yes. The line bundle $\mathcal{O}(d)$, with $d > 0$, is generated by the degree $d$ monomials of which there are $\binom{n+d}{d}$. So the number of generators you need grows with $d$. Your second question is a bit too naive. In general, if you have a morphism $X\to \mathbb{P}^{m}$, there is no link between the ...

1

A Cartier or $\mathbf Q$-Cartier divisor $D$ on a projective variety has well-defined interesection numbers $D \cdot C$ with curves $C$. If all those intersection numbers are $\geq 0$, we say $D$ is nef. But if $D$ is not $\mathbf Q$-Cartier, there is no way in general to define $D⋅C$, and so it is not meaningful to ask if $D⋅C \geq 0$ for all curves C.

1

The one-point compactification of $\mathbb{R}^2$ is a quotient of $\mathbb{P}^2(\mathbb{R})$, that is, smaller than $\mathbb{P}^2(\mathbb{R})$. To get $\mathbb{P}^2(\mathbb{R})$ you add to $\mathbb{R}^2$ the line at infinity which is in fact a projective line, $\mathbb{P}^1(\mathbb{R})$ ( homeomorphic to a circle). As a side note, there are many ...

1

Well, if you really want to work with Whitehead's axioms, here are the axioms as he put them in The axioms of projective geometry page 7: As you can see, axiom $II$ asserts there is a point, and axiom $III$ asserts there is a point other than the first point you picked. Then axioms $IV$ and $VII$ give you that the point you picked lies on a line. As ...

1

Let $A \in SU(3)$, $A=\matrix{(x_1& x_2&x_3)}$ where $x_1 \in \mathbb{C}^3$, $x_2 \in \mathbb{C}^3$, $x_3 \in \mathbb{C}^3$, with $\|x_1\|=\|x_2\|=\|x_3\|=1$ and $\langle x_1,x_2\rangle=\langle x_2,x_3 \rangle =\langle x_1,x_3 \rangle=0$. $U(2)$ acts on the first two vectors of $A$. If $B \in U(2)$, $\left(\matrix{x_1 & x_2}\right)B$ describes ...

1

A comprehensive presentation of Plücker coordinates and the corresponding embedding on an advanced level is in Griffiths, Harris: Principles of Algebraic Geometry. A more elementary version is explained in chapter 8 of Cox, Little, O'Shea: Ideals, Varieties, and Algorithms.

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