Complex and Conformal structure on a trivial tangent bundle of a higher dimensional manifold Let $M$ be an even dimensional Reimannian  manifold with trivial tangent bundle $TM$. Then there is a global orthonormal basis $\{e_1,e_2,...,e_{2n-1},e_{2n}\}$ of $TM$.
Define a bundle map $J :TM  \to TM$ which sends each $e_{2k-1}$ to $e_{2k}$ and 
 $e_{2k}$ to $-e_{2k-1}$ for $k=1,...,n$. Then, does this define a complex structure on $M$? Is it well-defined?
Ref
Complex structure v.s. conformal structure in more than 1 complex dimension
Proof of equivalence of conformal and complex structures on a Riemann surface.

Edit for understanding the answer.
First, we show that $M = (S^2\times S^2)\#(S^1\times S^3)\#(S^1\times S^3)$  has a trivial tangent bundle and to do so, it suffices to show that
(I)  the Euler numbr of  $M$  is zero and 
(II)  $M$ is a spin manifold.
Consider the (I).
For any triangularizable toplogical space $X$ and $Y$, there are simple algebraic relations between their Euler numbers.
$\chi(X \#Y) = \chi(X ) +\chi( Y)  -2,$
$\chi(X \times Y) = \chi(X ) \times \chi( Y). $
Using these formulas, we can calculate the Euler number of the manifold  $M = (S^2\times S^2)\#(S^1\times S^3)\#(S^1\times S^3)$ as follows.
$\chi(M) = \chi (S^2)\chi( S^2)+\chi(S^1)\chi( S^3)+\chi(S^1)\chi( S^3) -4.$
Because $\chi (S^2)=2$ and $\chi (S^1)=0$ and $\chi (S^3)=1-0+0-1=0$, we obtain
$\chi(M) = 2 \times 2 +0 \times \chi( S^3)+0 \times \chi( S^3) -4=0.$
Thus, $\chi(M)=0$ and thus (I) is O.K.
Next, consider the (II).
According to Albanese's comment, the connected sum of spin manifolds is spin.
Thus we show that the spaces $S^2\times S^2$, $S^1\times S^3$ and $S^1\times S^3$ are spin. To poove this, for example, we show $w_2(S^1\times S^3)=0$.
According to problem 4-A of page 54 in Milnor-Stasheff, 
$w_2(S^1\times S^3)=w_2(S^1) \times  w_0(S^3) + w_1(S^1) \times  w_1(S^3)   + w_0(S^1) \times  w_2(S^3) =0$
where we use the total Stiefel-Whitney class of sphere is 1, namely, $w(s^d)=1$ (page 42 example 1 in Milnor Stasheff).
 A: First of all, the fact that $M$ is Riemannian plays no role here. 
The endomorphism $J$ is indeed well-defined and as $J\circ J = -\operatorname{id}_{TM}$, it is an almost complex structure on $M$. However, it need not be an integrable almost complex structure. That is, it may not be an almost complex structure induced by a complex manifold. For example, $M = (S^2\times S^2)\#(S^1\times S^3)\#(S^1\times S^3)$ has trivial tangent bundle and hence has an almost complex structure as you describe, but it does not admit a complex structure.
In real dimension two, every almost complex structure is integrable, see this question. In real dimension four and above, there are examples of almost complex structures which are not integrable. In real dimension four, there are manifolds (such as $M$) which admit almost complex structures, none of which are integrable. It is an open problem whether such manifolds exist in real dimension six and above.
A: I would estimate yes in the sense each 2n is mapped one-to-one in the sense each +2n may be mapped onto the pure reals for the real part of the complex mapping and each of the -2n may be mapped onto the pure imaginary numbers of the imaginary part of the complex system.  I am sorry here if will not do.  I am an economist, not a mathematician.  Sincerely,  Richard Anthony Baum 
