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

Question

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 $\mathbb{CP}^2$?

Answer 1

• 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 Moebius 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 $PGL(2,\mathbb{CP}^n)$.

Answer 2

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.

Answer 3

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 $Pol(P_2)$.

Answer 4

Anirbit,

this is to answer your last question and your refined question in the comment to JC:

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.