# Is there a four-manifold whose tangent bundle is an endomorphism bundle?

Is there a smooth four-manifold $M$ such that $TM \cong \operatorname{End}(E)$ for some rank $2$ bundle $E \to M$?

If $M$ is parallelisable, then one can take $E$ to be the trivial rank $2$ bundle over $M$. I would like to know if there is a non-trivial example.

As sections of the bundle $\operatorname{End}(E)$ are vector bundle morphisms $E \to E$ (covering the identity map $M \to M$), the identity map $\operatorname{id}_E$ defines a nowhere-zero section of $\operatorname{End}(E)$. Therefore, such an $M$ must have Euler characteristic zero by the Poincaré-Hopf Theorem.

Using the real splitting principle, I was able to show that $w(\operatorname{End}(E)) = 1 + w_1(E)^2$. In particular, $M$ must be orientable and $w_2(M)$ is a square; such manifolds were asked about here, but no explicit four-manifold examples were given.

Choosing an orientation for $M$, I would like to compute its first Pontryagin class and hence the signature of $M$, but I have been unable to do so. Even if I had this information, I would have no idea how to proceed.

Added Later: Thanks to Igor Belegradek for his help. Using his advice, one can show that $p_1(\operatorname{End}(E)) = -4c_2(E_{\mathbb{C}}) = 4p_1(E)$ where $E_{\mathbb{C}} = E\otimes_{\mathbb{R}}\mathbb{C}$ is the complexification of $E$. If $E$ is orientable, it can be viewed as a complex line bundle, in which case $E_{\mathbb{C}} = E\oplus\overline{E}$ and the expression for $p_1(\operatorname{End}(E))$ reduces to $4c_1(E)^2$ as in my answer below.

This expression for $p_1$, together with the Dold-Whitney Theorem, might also lead to examples where $E$ is non-orientable.

• Any rank 4 bundle over a closed oriented 4-manifold is uniquely determined by $w_2, w_4$, and $p_1$, see p.647 of [Dold-Whitney, Classification of Oriented Sphere Bundles Over A 4-Complex]. You can compute the Chern classes of the complexification in the same way as the SW classes, see e.g. math.stackexchange.com/questions/989147/…. This will give you $p_1$. I do not want to complete the computation because this is a good exercise, hence not posting this as an answer. Also $End(\xi)$ is isomorphic to $\xi\otimes\xi$. – Igor Belegradek Apr 18 '17 at 21:09
• @IgorBelegradek: I believe Dold-Whitney is for oriented bundles, and the classes are $w_2$, $e$, and $p_1$ (otherwise $TS^4$ and $\varepsilon^4$ would be a counterexample). I haven't been able to make any progress with calculating $p_1$ in general using the splitting principle. If I assume $E$ is orientable, then I can calculate $p_1$, but not via the splitting principle. I will include what I have done later today. – Michael Albanese Apr 21 '17 at 16:06
• You said above that $M$ must be orientable, which is why I thought that $w_2$. $w_4$, $p_1$ suffice. Dold-Whitney also include results for non-orientable bundles but then the answers are more complicated. As for $p_1$ why cannot you compute $c_2$ of the complexifcation in the same way you computed $w_2$, or alternatively, by the formulas I linked above? – Igor Belegradek Apr 21 '17 at 16:51
• @IgorBelegradek: I still don't think $w_2$, $w_4$ and $p_1$ suffice (they are all stable classes, they can't tell the difference between stably trivial and trivial). The formulas you linked to above tell you how to obtain the Chern classes of a bundle formed out of complex vector bundles. Here $E$ is not assumed to be complex (it isn't even assumed to be orientable). In the computation of the Stiefel-Whitney classes I could split $E$ as a sum of real line bundles and get a corresponding splitting of $\operatorname{End}(E)$. What is the analogue of this here given that $E$ may not be complex? – Michael Albanese Apr 21 '17 at 18:24
• I suggest read the relevant portion of Dold-Whitney, e.g. at see maths.ed.ac.uk/~aar/papers/doldwhit.pdf, and specifically, the corollary on p.674. The 4th cohomology of the base has no torsion because the manifold $M$ is orientable (as you claim in the question, i.e., you say that $TM=End(E)$ is orientable). Another thing I suggest is that you read the definition of Ponryagin classes, e.g. in Milnor-Stasheff. If you did that, you would know that $p_1(\xi)$ is defined as $-c_2(\xi\otimes\mathbb C)$. Orientability of $\xi$ is not needed. – Igor Belegradek Apr 21 '17 at 19:17

Note, this answer was conceived before I understood Igor Belegradek's comments regarding the calculation of $p_1$.

Suppose $E$ is orientable, i.e. $w_1(E) = 0$. Then $w_2(\operatorname{End}(E)) = w_1(E)^2 = 0$, so $M$ must be spin.

Choosing an orientation for $E$, we can view $E$ as a complex line bundle. Then

$$\operatorname{End}(E) = \operatorname{End}_{\mathbb{C}}(E)\oplus\overline{\operatorname{End}}_{\mathbb{C}}(E)$$

where the terms of the decomposition are complex linear and complex antilinear endomorphisms respectively. If $J$ denotes the almost complex structure on $E$, then the decomposition is given by $L \mapsto \frac{1}{2}(L - JLJ) + \frac{1}{2}(L+JLJ)$. Note that $\operatorname{id}_E$ defines a nowhere-zero section of $\operatorname{End}_{\mathbb{C}}(E)$, so $\operatorname{End}_{\mathbb{C}}(E) \cong \varepsilon_{\mathbb{C}}^1$ (alternatively, $\operatorname{End}_{\mathbb{C}}(E) \cong E^*\otimes E \cong \varepsilon^1_{\mathbb{C}}$). On the other hand, a complex anti-linear endomorphism of $E$ can be viewed as a complex linear homomorphism $E \to \overline{E}$, so

$$\overline{\operatorname{End}}_{\mathbb{C}}(E) \cong \operatorname{Hom}_{\mathbb{C}}(E, \overline{E}) \cong E^*\otimes\overline{E} \cong \overline{E}^2.$$

Therefore

\begin{align*} p_1(\operatorname{End}(E)) &= p_1(\varepsilon_{\mathbb{C}}^1\oplus\overline{E}^2)\\ &= p_1(\overline{E}^2)\\ &= -c_2(\overline{E}^2\otimes_{\mathbb{R}}\mathbb{C})\\ &= -c_2(\overline{E}^2\oplus E^2)\\ &= -c_1(\overline{E}^2)c_1(E^2)\\ &= -4c_1(\overline{E})c_1(E)\\ &= 4c_1(E)^2. \end{align*}

Now we can use the Dold-Whitney Theorem.

Dold-Whitney Theorem: Oriented rank four real bundles on an closed, orientable four-manifold are uniquely determined by their second Stiefel-Whitney class $w_2$, Euler class $e$, and first Pontryagin class $p_1$.

Under the assumption that $E$ is orientable, the bundle $\operatorname{End}(E)$ is an oriented rank four real bundle on an orientable four-manifold with $w_2(\operatorname{End}(E)) = 0$, $e(\operatorname{End}(E)) = 0$ and $p_1(\operatorname{End}(E)) = 4c_1(E)^2$. If we can find an orientable four-manifold $M$ which is spin, has $\chi(M) = 0$, and $p_1(M) = 4c_1(E)^2$ for some complex line bundle $E$ on $M$, then by the Dold-Whitney Theorem, $TM \cong \operatorname{End}(E)$.

Let $M$ be a closed, smooth, spin four-manifold with $\chi(M) = 0$. If $\tau(M) = 0$, then $p_1(M) = 0 = 4c_1(E)^2$ where $E$ is the trivial complex line bundle. If $\tau(M) \neq 0$, then the intersection form of $M$ is of the form $\pm 2mE_8\oplus nH$ for some non-negative integers $m, n$. By Donaldson's theorem on definite intersection forms, if $m \neq 0$, then $n \neq 0$. So if $\tau(M) \neq 0$, it must have a copy of the hyperbolic lattice $H$ in its intersection form. Therefore every even integer is of the form $c^2$ for some $c \in H^2(M; \mathbb{Z})$. By Rokhlin's theorem, $\tau(M)$ is a multiple of $16$, so $\frac{3}{4}\tau(M) = \frac{1}{4}p_1(M)$ is a multiple of 12 (in particular, even), so there is a class $c$ such that $c^2 = \frac{1}{4}p_1(M)$, i.e. $p_1(M) = 4c^2$. As every element of $H^2(M; \mathbb{Z})$ is the first Chern class of some complex line bundle, we see that $p_1(M) = 4c_1(E)^2$ for some complex line bundle $E$. Therefore, we have the following:

Let $M$ be a closed, smooth, spin four-manifold with $\chi(M) = 0$. Then $TM \cong \operatorname{End}(E)$ for some $E$.

Moreover, by Dold-Whitney, such an $M$ is parallelisable if and only if $\tau(M) = 0$. In particular, if $M$ is spin with $\chi(M) = 0$, $\tau(M) \neq 0$, we get a non-trivial example.

Here's one way to construct such examples. Let $X$ be a closed, smooth, simply connected, spin four-manifold and set $M = X\# k(S^1\times S^3)$ where $k = \frac{1}{2}\chi(X)$. Then $M$ is a spin manifold with $\chi(M) = 0$, so $TM \cong \operatorname{End}(E)$ for some $E$. As $\tau(M) = \tau(X)$, we obtain non-trivial examples whenever $\tau(X) \neq 0$. The simplest non-parallelisable manifold that arises from this construction is

$$M = K3\#12(S^1\times S^3).$$

In this case $E$ is the unique complex line bundle with $c_1(E)^2 = -12$.