# $Aut(\mathbb{CP}^n)$ [..especially $n=1$ and $n=2$..]

I am confused and curious about the meaning of the $Aut(\mathbb{CP}^n)$.

• Is what is called the "linear automorphism group" of $\mathbb{CP}^n$ the same as $Aut(\mathbb{CP}^n)$? It somehow seems to me to be very non-trivial if they are the same things.

• I see the statement that $Aut(\mathbb{CP}^1) = { z \mapsto \frac{az+b}{cz+d} , ad-bc \neq 0 }$ How am I supposed to interprete this statement? If $z$ is the homogeneous coordinate then its not clear to me that this map is well defined on a projective space. How does one prove this?

• Is there a similar way to write down $Aut(\mathbb{CP}^2)$?

• One wants to show that any two irreducible conic sections in $\mathbb{CP}^2$ are "projectively equivalent". I would like to know how this is shown. Does this mean that there exists an element of $Aut(\mathbb{CP}^2)$ that transforms one to the other? Is there a way to write down a general expression for an irreducible conic in $\mathbb{CP}^2$?

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$Aut(\mathbb{CP}^n)$ is just $PGL(n,\mathbb{C})$. –  J.C. Ottem Sep 14 '11 at 20:43
See Griffiths-Harris, chapter 1 –  algori Sep 14 '11 at 20:49
@Anirbit: Of course the answer to this question depends on the structures on $\mathbb{CP}^n$ you'd like to preserve. Based on your last question, it seems like you'd like automorphisms as an algebraic variety, which is indeed $PGL(n, \mathbb{C})$. But if you view $\mathbb{CP}^n$ as "just" a topological space or manifold, it has many more automorphisms. –  Daniel Litt Sep 14 '11 at 21:55
@Daniel Can you kindly clarify what is the defintion you are using of "automorphisms as an algebraic variety"? I guess the statement is that all that is isomorphic to the "linear automorphisms" (ones which come down from the vector space automorphisms of $\mathbb{C}^{n+1}$) –  Anirbit Sep 17 '11 at 18:34

• The group $Aut(\mathbb{P}^n)$ is just $PGL(n,\mathbb{C})$. To see why, consider any automorphism $\sigma$ of $\mathbb{P}^n$. You can easily show that $\sigma^*O(1)=O(1)$, since $\sigma$ induces an automorphism of the Picard group and $\sigma^*O(1)$ is effective. In particular, $\sigma$ induces an automorphism on $V=H^0(\mathbb{P}^n,O(1))$ which is an $(n+1)$-dimensional vector space. Now by the correspondence between sections of very ample line bundles and projective embeddings we see that $\sigma$ is actually determined by the automorphism on $V$ up to a scalar. Hence $\sigma$ comes from $PGL(n,\mathbb{C})$.

• You identify the mobious transformation $\frac{az+b}{cz+d}$ with the element $\begin{pmatrix} a & b\cr c & d\end{pmatrix}\in PGL(2,\mathbb{C}^2)$ for this correspondence. This corresponds restricing the automorphism to the affine chart with cordinate $z$.

• No, there are no analogous Moebious transformations for $n\ge 2$.

• All irreducible plane conics (which corresponds to degree 2 forms in $x_0,x_1,x_2$) are isomorphic to $x_0^2+x_1^2+x_2^2=0$, as you can easily check by a change of variables (i.e., an element of $PGL(2,\mathbb{CP}^n)$.

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@J.C.Ottem Thanks for your answer. What I was asking about the $Aut\mathbb{CP}^1$ is how do I see that two points of $\mathbb{CP}^1$ which are non-zero scalar multiples of each other are being mapped to the same image point. And for the last part are you saying that I can take the matrix of coefficients corresponding to the homogeneous plane conic and show that it has an eigen value of $1$ with a multiplicity of $3$? How does one force the irreducibility condition? –  Anirbit Sep 15 '11 at 2:32
@JC: in your last bullet point you obviously meant irreducible conic, but perhaps you should add it, so everyone knows what you mean. Cheers! :) –  Sándor Kovács Sep 15 '11 at 6:13
That's right, thanks! –  J.C. Ottem Sep 15 '11 at 8:52
This is a fine answer if you are already happy with sheaves and the Picard group, but it would be interesting to know if there was a more elementary one. –  Neil Strickland Sep 15 '11 at 12:06
Yes, that is interesting. Perhaps one could even hope that the above proof could be translated into a more elementary one. After all, line bundles and the Picard group are not very complicated objects in the case of $\mathbb{P}^n$. –  J.C. Ottem Sep 15 '11 at 12:44

You are right. This is indeed a non-trivial fact: $\mathbb P^n$ only has linear automorphisms. Here is a sketch of the proof: Let $\alpha$ be an automorphism. Then $\alpha^*$ is an automorphism of $\mathrm{Pic}\ \mathbb P^n\simeq \mathbb Z$, but then $\mathscr O(1)$ is taken to either itself or to $\mathscr O(-1)$. But the second is actually impossible since $\mathscr O(1)$ has non-zero global sections while $\mathscr O(-1)$ does not. Therefore every automorphism fixes $\mathscr O(1)$ and hence they are linear.

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This should be a comment, except that it is too long.

I like to recall a classical way (not involving eigenvalues) of showing that every smooth plane conic over a field of characteristic $\ne 2$ can be put in diagonal form.

Assume that the irreducible conic $Q$ is defined by a $3\times 3$ symmetric matrix $A$; $A$ has rank $3$, since it $Q$ is irreducible, and is determined uniquely up to multiplying by a nonzero scalar. To a point $P=[v]$ in the plane one can associate the {\it polar line} $Pol(P)$, which is the line of ${\mathbb P}^2$ defined by $vA^tx=0$, where $x:=[x_0, x_1, x_2]$. The map $P\mapsto Pol(P)$ is a projective isomorphism ${\mathbb P}^2\to ({\mathbb P}^2)^*$. (The polar can also be defined geometrically as follows: if $P\in Q$, then $Pol(P)$ is the tangent to $Q$ at $P$, while if $P\notin Q$, then there are two lines passing through $P$ that are tangent to $Q$ at points $R_1$ and $R_2$ and $Pol(P)$ is the line joining $R_1$ and $R_2$).

Three distinct points $P_1,P_2,P_3$ in the plane are an {\it autopolar triangle} if $Pol(P_i)$ is the line joining $P_j$ and $P_k$, where $i,j,k$ are any permutation of $1,2,3$. It is easy to check that $Q$ is in diagonal form in a system of homogeneous coordinates iff the three coordinate points of the system are an autopolar triangle. So it is enough to show that an autopolar triangle exist and this is easily done as follows: pick a point $P_1\notin Q$, pick $P_2\in Pol(P_1)\setminus Q$ and define $P_3$ to be the intersection point of $Pol(P_1)$ and $Pol(P_2)$.

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Anirbit,

Forget about eigenvalues, their explicit value gets lost in the projective world and the only thing that matters is whether they are zero or not. $$[x_0:x_1:x_2]\mapsto [\lambda_0x_0:\lambda_1x_1:\lambda_2x_2]$$ with $\lambda_0\lambda_1\lambda_2\neq 0$ is an automorphism of $\mathbb P^2$, so in order to prove that any irreducible conic is isomorphic to $x_0^2+x_1^2+x_2^2=0$ you only need to notice that the matrix defining your conic is symmetric and hence diagonalizable. The irreducibility condition translates to the matrix being of full rank, i.e., invertible, i.e., having non-zero eigenvalues, so using the above automorphism you can get the form $x_0^2+x_1^2+x_2^2=0$.

To see why the statement about irreducubility is true notice that the diagonalizable part does not use anything about irreducibility, so you get something like $\lambda_0x_0^2+\lambda_1x_1^2+\lambda_2x_2^2=0$, which is the equation of two lines (possibly a double line) if and only if $\lambda_0\lambda_1\lambda_2=0$.

There is also a heuristic proof of this: the matrix corresponding to the quadric which is the union of two lines is (sort of) the tensor product of a column matrix with a row matrix and hence cannot be invertible. This is not quite solid reasoning, but it gives you an idea.

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@Sandor Thanks for the reply. I am confused about something. I get your point about finding a basis where the conic reduces to a sum of 3 coordinate squares. But the similarity transformation that will take one conic to this form will not work for the other conic. In general I will get for one conic an expression like $x_0^2 + x_1^2 + x_2 ^2 = 0$ and for the other I will get some $x'_0 ^2 + x'_1^2 + x'_2^2=0$ and there will be some non-trivial linear transformation connecting the $x'$ with $x$. –  Anirbit Sep 17 '11 at 17:44
@Sandor How does one see from here that these two conics are connected by some element of $Aut(\mathbb{CP}2)?I don′t know how to imagine/write downAut(\mathbb{CP}^2)!It should be the GL(3,\mathbb{C})$ modded out by all diagonal matrices? I guess one has to show that if $A$ and $B$ are two matrices in $GL(3,\mathbb{C})$ act on some element of $\mathbb{C}^3$ and take it to two other points which are scalar multiples of each other then $A$ and $B$ are also scalar multiples of each other...but I am not sure how to set this up. –  Anirbit Sep 17 '11 at 17:50
@Anirbit: 1) For your first comment: the $x$'s and the $x'$'s are connected by a homogenous linear transformation. That will give you an element of $\mathrm{Aut}\,\mathbb P^2$ 2) For the second. In my first answer I showed why there are only linear transformations. Those are indeed the ones coming from $\mathrm{GL}(3,\mathbb C)$ but the diagonal matrices act trivially. So, yes, it should be PGL. I am not sure what you mean by "how to set this up" –  Sándor Kovács Sep 18 '11 at 6:27
@Sandor I can see that there is an linear transformation which connects the two different diagonalizing basis but what is not clear to me is that it transforms the zero-set of the either to the other. That seems to need to demand something special about the initial two similarity transformation that diagonalized each of them.(like may be unitarity). Or more elementarily isn't coefficient matrix complex? Then isn't it true that a complex symmetric matrix may not even be diagonalizable? Kindly explain. –  Anirbit Sep 19 '11 at 7:23
@Sandor And what equation do you have in mind for a line in C^3? I was trying to understand your comment about the product of eigenvalues being non-zero. –  Anirbit Sep 19 '11 at 7:27