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When looking at a projective variety, we can intersect the variety with the standard affine patches and the union of these intersections give the projective variety. So the variety can be viewed as a projective variety over $\Bbb P^n$ or as an affine variety called the "affine cone" over $\Bbb C^{n+1}$,but with the points on the same line identified. Am I right?

But I am still not able to understand what an affine cone is.

I do not know about vector bundles or sheaves. I only have a very basic background knowledge in the subject.

So kindly explain what an affine cone is, in simple terms.

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    $\begingroup$ "So kindly explain what an affine cone is ,in simple terms." I think you gave the correct definition in your question, so I'm not sure what it is you are asking exactly... it is called "affine" because it lives in $\mathbb{C}^{n+1}$ which is an affine space, and it's called a "cone" because it has the property that if a point $(x_0, \ldots, x_{n+1})$ belongs to it then so do the points $t (x_0, \ldots, x_{n+1})$ for any $t \in \mathbb{C}$ $\endgroup$ – Glougloubarbaki Jun 1 '17 at 15:24
  • $\begingroup$ What I don't understand is how different the two perceptions are . Also, How are they useful? Can someone give me examples wherein, looking at it as an affine cone is more beneficial than looking at it as a projective variety? $\endgroup$ – Nivedita Jun 1 '17 at 15:39
  • $\begingroup$ it is strictly and exactly the same thing. sometimes if you want to work in coordinates and make explicit computations, it might be more convenient to use the cone, but the more intrinsic object is the projective variety in $\mathbb{P}^n$. but again, it is the same thing $\endgroup$ – Glougloubarbaki Jun 1 '17 at 15:48
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    $\begingroup$ think of this: do you prefer to talk about periodic functions of period $2\pi$ on $\mathbb{R}$ or to talk of functions defined on the circle? it is the same thing. $\endgroup$ – Glougloubarbaki Jun 1 '17 at 15:51
  • $\begingroup$ Ahhh! I get it now! Thank you! $\endgroup$ – Nivedita Jun 1 '17 at 15:54
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Let's consider an explicit example. Look at the equation $xy=z^2$ in the projective plane $\mathbb{P}^2$ with coordinates $[x:y:z]$. The given locus is a quadric, i.e. a curve isomorphic to $\mathbb{P}^1$ and with the property that its intersection with a line (a copy of $\mathbb{P}^1$ given by linear equations) is two points (including the case of one point with mutliplicity 2).

Now, the way way we build the cone is the following. Remember that $\mathbb{P}^2$ with coordinates $[x:y:z]$ is obtained from $\mathbb{A}^3$ with coordinates $(x,y,z)$, removing $(0,0,0)$, and quotienting by the rescaling action of the group of units of the ground field $k^*$. In particular, the equations that define our quadric (or, more generally, the projective variety in $\mathbb{P}^n$ you start with) still make sense in $\mathbb{A}^3$ ($\mathbb{A}^{n+1}$ respectively). Some algebra computations show that the locus you obtain has one extra dimension than what you started with. The reason is, as you correctly wrote, that we are taking the space of lines over the projective variety.

It is important to stress that we are not considering these lines as points in the projective space, but as honest lines in affine space. Thus, the picture that the real points (i.e. the points that live over $\mathbb{R}$) of the above example are the following: you can think of the projective conic as a cricle, and the cone over it is the honest right cone with circular base.

Thus, the affine cone over a projective variety is a cone whose ''horizontal slices'' recover the variety you started with. Notice that the cone is always singular at the origin.

By construction, the geometry and the properties of these two objects are closely related. For instance, the cone is a normal variety if and only if the embedding of the projective variety is projectively normal.

Cones are also very important, since they provide a nice list of examples of computable singular varieties, where one can test ideas and computations about singular varieties.

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  • $\begingroup$ Thank you so much! I have understood the answer to a large extent but there is something I am not very clear about. How is the locus of the above equation a quadric that is isomorphic to P1? $\endgroup$ – Nivedita Jun 1 '17 at 16:50
  • $\begingroup$ @Nivedita Take $\mathbb{P}^1$ with coordinates $[t_0:t_1]$ and define the embedding $\varphi: \mathbb{P}^1 \rightarrow \mathbb{P}^2$ by $x=t_0^2$, $y=t_1^2$, $z=t_0 t_1$. $\endgroup$ – Stefano Jun 1 '17 at 16:54
  • $\begingroup$ @Nivedita This, together with the characterisation of quadrics by rank of the matrix that represents them shows you that all non-singular quadrics in $\mathbb{P}^2$ (provided that the ground field is algebraically closed) are isomorphic to $\mathbb{P}^1$. $\endgroup$ – Stefano Jun 1 '17 at 16:57
  • $\begingroup$ Perfect!! One more question: You have mentioned that some algebra computations show that the locus we obtain has one extra dimension than what we start with. We start with the quadric xy=z^2, and homogenisation would give the same equation. So how is the increase in the dimension justified? $\endgroup$ – Nivedita Jun 1 '17 at 17:00
  • $\begingroup$ When you look at it in $\mathbb{A}^3$, you look at the ring $\mathbb{C}[x,y,z]/(xy-z^2)$. So you quotient a 3-dimensional ring by an irreducible element, you get a 2-dimensional ring (a chain of primes could be $(xy-z^2) \subset (xy-z^2,z-1) \subset (xy-z^2,z-1,x-1)$). When you look at it in $\mathbb{P}^2$ with coordinates $[x:y:z]$, you look at what you get on an affine cover. The usual cover is given by the open sets $\lbrace x \neq 0 \rbrace$, $\lbrace y \neq 0 \rbrace$ and $\lbrace z \neq 0 \rbrace$. Consider the first one. It is equivalent to set $x=1$. $\endgroup$ – Stefano Jun 1 '17 at 17:06

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