Question regarding the Taub-NUT metric on $\mathbb{S}^3 \times \mathbb{R}^+$. I have a question regarding Claude Lebrun's paper Complete Ricci-flat Kähler metrics on $\mathbb{C}^n$ need not be flat. In the introduction of the paper, he writes that the Taub-NUT metric is given explicitly by $$g = \frac{\rho +1}{4 \rho} d\rho^2 + \rho(1+\rho) \left[ \sigma_1^2 + \sigma_2^2 \right] + \frac{\rho}{\rho+1}\sigma_3^2,$$ where $\sigma_1, \sigma_2, \sigma_3$ is a left-invariant coframe for $\mathbb{S}^3$ and where $\rho \in \mathbb{R}^+$. He continues to mention that there is a common erroneous assertion which maintains that this metric is not Kähler -- the reasoning being that with respect to the most obvious integrable almost complex structure $(\sigma_1 \mapsto \sigma_2, \sigma_3 \mapsto -(1+\rho)d\rho/2\rho)$, the metric is Hermitian but not Kähler. However, the metric has self-dual curvature tensor, and so has holonomy $SU(2)$. 

Q: I want to verify that the above almost complex structure is integrable. 

It seems that the most straightforward way of doing this is by showing that the Nijenhuis tensor vanishes identically. We can write vector fields $X$ and $Y$ on $\mathbb{S}^3 \times \mathbb{R}^+$ as $X = \alpha_1 \sigma_1 + \alpha_2 \sigma_2 + \alpha_3 \sigma_3 + \alpha_4 d\rho$ and $Y = \beta_1 \sigma_1 + \beta_2 \sigma_2 + \beta_3 \sigma_3 + \beta_4 d\rho$. Then \begin{eqnarray*}
JX &=& \alpha_1 \sigma_2 - \alpha_2 \sigma_1 - \alpha_3 \frac{1+\rho}{2\rho} d\rho - \alpha_4 \frac{2\rho}{1+\rho} \sigma_3, \\
JY &=& \beta_1 \sigma_2 - \beta_2 \sigma_1 - \beta_3 \frac{1+\rho}{2\rho} d\rho - \beta_4 \frac{2\rho}{1+\rho} \sigma_3. 
\end{eqnarray*}
The result Lie bracket computations become very cumbersome to treat. Is there an easier way of showing that the above almost complex structure is integrable? 
 A: A necessary and sufficient condition for the integrability of $J$ is that the exterior differential of a $(1,0)$-form (with respect to $J$) is a $(2,0)$-form plus a $(1,1)$-form (and thus has no $(0,2)$ component).
I think that $d\sigma_1 = -\sigma_2 \wedge \sigma_3$ (please double check the sign) and cyclically.
So $d(\sigma_1 + i \sigma_2) = -\sigma_2 \wedge \sigma_3 - i \sigma_3 \wedge \sigma_1 = (-\sigma_2 + i \sigma_1) \wedge \sigma_3 = i(\sigma_1 + i\sigma_2) \wedge \sigma_3$.
But then $\sigma_1 + i \sigma_2$ is a $(1,0)$-form, while $\sigma_3$ has a $(1,0)$-part and a $(0,1)$-part, so the right-hand side of the previous formula is the sum of a $(2,0)$-form and a $(1,1)$-form, as required.
It remains only to do a similar calculation for
$$d\left(\sigma_3 - i(1+\rho)\frac{d\rho}{2 \rho}\right).$$
Note that the second term inside the parentheses is of the form $f(\rho) d\rho$, so its exterior differential is thus $f'(\rho) d\rho \wedge d\rho = 0$.
On the other hand, $d \sigma_3 = - \sigma_1 \wedge \sigma_2$, so we get:
$$d\left(\sigma_3 - i(1+\rho)\frac{d\rho}{2 \rho}\right) = -\sigma_1 \wedge \sigma_2 = -\frac{i}{2}(\sigma_1 + i \sigma_2) \wedge (\sigma_1 - i \sigma_2),$$
which is a $(1,1)$-form, with respect to $J$. Hence $J$ is integrable.
