1
$\begingroup$

I am looking at page 15 here, starting from the line 'Let us suppose we are given a...' The Zoll projective structure is not relevant to this question, so don't worry about that.

We take, in this example, $M=\mathbb{RP}^2$, which is a manifold of (real) dimension 2. Hence the tangent bundle $TM$ is of (real) dimension 4. Complexifying this, we get $T_\mathbb{C}M=\mathbb{C}\otimes{TM}$ of complex dimension 4. Projectivising this, we get:

\begin{equation} \mathbb{P}T_\mathbb{C}M = (\mathbb{C}\otimes{TM}-\{0\})/{\mathbb{C}^\times} \end{equation}

Since $\mathbb{C}\otimes{TM}$ is of complex dimension 4, projectivising it gives us $\mathbb{P}T_\mathbb{C}M$ of complex dimension 3. Hence $\mathbb{P}T_\mathbb{C}M$ also has the structure of a real manifold of dimension 6.

However, a couple of lines below in the paper, $\mathbb{P}T_\mathbb{C}M$ is stated as being a manifold of (real) dimension 4. Where have I gone wrong?

$\endgroup$
  • 1
    $\begingroup$ $TM$ is a four dimensional manifold, but it is just a 2 dimensional bundle over $M$. So $\mathbb C \otimes TM$ is of complex dimension 2 $\endgroup$ – user99914 Jul 12 '16 at 11:12
4
$\begingroup$

You're conflating vector bundles with their total spaces in a way which is confusing you; I recommend that you firmly distinguish them in the future. The complexification of the tangent bundle of $M = \mathbb{RP}^2$ is a complex vector bundle of rank $2$, hence its underlying real vector bundle is a real vector bundle of rank $4$. This means that its total space has dimension $4 + 2 = 6$, and after dividing by the $\mathbb{C}^{\times}$ action the dimension is $6 - 2 = 4$.

In your calculation you inappropriately doubled the dimension of $M$ itself, which is in no sense being complexified here. The dimension of the total space of a vector bundle is the sum of the dimension of the base space and the rank of the vector bundle, which are in general unrelated; that's one reason why it's confusing to conflate the vector bundle and its total space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.