Does $S^1 \times S^3$ admit a complex structure?

Question

Does $S^1 \times S^3$ admit a complex structure?

It is parallelizable, so it admits an almost complex structure...is there a nice way to see that this is or is not integrable?

Answer 1

Which almost complex structure do you have in mind? In fact they are better than paralelizable ($S^1$ and $S^2$), their tangen bundles are Lie algebras so checking Nijenhuis tensor is very easy (it is linear problem) whatever structure given by that paralelization by fundamental vector fields you have in mind. It is even better, since both $S^1$ and $S^3$ are compact Lie groups you can find a left invariant complex structure on $S^1 \times S^3$ so Yes it admit complex structure.

EDIT: The structure you are indicating is not vell defined since you have to choose an identification of tangent space with $\mathbb{R}^4$. You would like to write something like that:

$\bullet$ $TS^1$ is paralelizable by a single vector field $\xi$ let say

$\bullet$ $TS^3$ is paralelizable by vector fields $e_1, e_2, e_3$

that vector fields $lifts$ to vector fields on $S^1 \times S^3$ and give a paralelization of $T(S^1 \times S^3)$ where Lie bracket between them is:

$\bullet$ $[\xi, e_i]=0, [e_1,e_2]=e_3, [e_2,e_3]=e_1, [e_3,e_1]=e_2.$

Now you can define an almost complex structure for example by:

$$e_2 \mapsto \xi \mapsto - e_2$$ $$e_1 \mapsto e_3 \mapsto -e_1.$$ That is an almost complex structure yet you have to chcecke if it is integrable - so if $$[J,J](X,Y)=0$$ for all $X,Y \in \{e_1,e_2, e_3, \xi\}$.