2
$\begingroup$

I can across the following lemma:

"There are maps $F_i : \mathbb{P}_{\mathbb{C}}^n \mapsto \mathbb{P}_{\mathbb{C}}^n$ such that for every $i = 1 \dots n $ $F_i \sim \text{Id}_{\mathbb{P}_{\mathbb{C}}^n}$ and $F_i (H_0)=H_i $, where $H_i := \{ \mathbf{x} \in \mathbb{P}_{\mathbb{C}}^n: x_i = 0 \}$. In particular,

$$ \eta_{H_0}=\eta_{H_i} $$

where $\eta_{(\cdot)}$ denotes the (closed) Poincarè dual of the embedded submanifold $(\cdot)$ in $\mathbb{P}_{\mathbb{C}}^n$."

Since I cannot directly find those maps, I tried a similar but slightly different approach to get (only) the final conclusion:

Let $F_i : \mathbb{P}_{\mathbb{C}}^n \mapsto \mathbb{P}_{\mathbb{C}}^n$ be the map such that $[x_0: \dots : x_n] \mapsto [x_0 + x_i : x_1 : \dots : x_n]$, clearly $F_i(H_0)=\{ x_0 = x_i \}$ and $F_i$ is (smoothly) homotopic to the identity through the map $[x_0: \dots : x_n] \mapsto [x_0 + tx_i : x_1 : \dots : x_n]$. In a similar fashion, define for $i \neq 0$ $G_i : \mathbb{P}_{\mathbb{C}}^n \mapsto \mathbb{P}_{\mathbb{C}}^n$ as $[x_0 : x_1 : \dots : x_n] \mapsto [x_0 : \dots \underbrace{x_0+x_i}_{i-th} \dots x_n]$, now we see that again $G_i \sim \text{Id}_{\mathbb{P}_{\mathbb{C}}^n }$ and $G_i(H_i)=\{ x_0=x_i$ }. Whence:

$$ \eta_{H_0}=\eta_{\{x_0=x_i\}}=\eta_{H_i} \qquad \forall i=1 \dots n $$

Where we used the fact that (closed) Poincarè duals are invariant under maps smoothly homotopic to the identity.

I was wondering if my proof is correct, furthermore, it seems like I got very close to find also those $F_i$s mentioned in the Lemma, but I tried to compose and modify my maps with little success: maybe someone is able to provide an insight on this one, thank you.

$\endgroup$

1 Answer 1

3
$\begingroup$

This is correct, except that $F_i(H_0)$ and $G_i(H_i)$ are $\{x_0=-x_i\}$, not $\{x_0=x_i\}$.

Here's a good way to think about a more direct approach. If you restrict yourself to maps $\mathbb{P}^n_\mathbb{C}\to\mathbb{P}^n_\mathbb{C}$ induced by linear maps $\mathbb{C}^{n+1}\to\mathbb{C}^{n+1}$ linear maps as you did, what you want is a path between the identity and the map that swaps the $0$th and $i$th coordinates on $\mathbb{C}^{n+1}$. Keeping the other $n-1$ coordinates fixed, it would suffice to have a path from $\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$ to $\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$ in $GL_2(\mathbb{C})$. Notice that this is not possible over $\mathbb{R}^2$, since the first matrix has determinant $1$ and the second has determinant $-1$ and so the determinant would have to pass through $0$. But over $\mathbb{C}$ we can go around $0$. Indeed, $\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$ is diagonalizable over $\mathbb{C}$ with eigenvalues $i$ and $-i$. We can then just choose paths from $i$ to $1$ and $-i$ to $1$ which do not pass through $0$, to get a path to the identity matrix.

Or, to more concretely build on what you have done, you found paths from $\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$ to $\begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}$ and from $\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$ to $\begin{pmatrix} 0 & 1 \\ 1 & 1\end{pmatrix}$. It would thus suffice to get a path from $\begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}$ to $\begin{pmatrix} 0 & 1 \\ 1 & 1\end{pmatrix}$. If you keep the bottom row fixed, this just means you want a path from $(1,0)$ to $(0,1)$ that does not pass through any scalar multiple of $(1,1)$ (so that every step will be an invertible matrix). This is easy to do when you can use complex numbers: for instance, you can take a path from $1$ to $0$ in the upper half-plane on the first coordinate and a path from $0$ to $1$ in the lower half-plane on the second coordinate.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .