Conormal sheaf in the analytic category and Kähler differentials As discussed here the Kähler differentials $\Omega^1_K$ and the ordinary differentials $\Omega^1$ of a complex manifold $(M,\mathcal{O}_M)$ are not the same. As sheaves of $\mathcal{O}_M$-modules, $\Omega^1$ is finitely generated while $\Omega^1_K$ is not.
Now, Kähler differentials may be constructed as follows:
Take $\mathcal{O}_M\otimes_{\mathbb{C}}\mathcal{O}_M\to \mathcal{O}_M$ the multiplication map and denote by $J$ its kernel. Then $\Omega^1_K\cong J/J^2$.
On the other hand, take the multiplication $\mathcal{O}_M\hat{\otimes}\mathcal{O}_M\to \mathcal{O}_M$ of the analytic tensor product ($\mathcal{O}_{\mathbb{C}^k}\hat{\otimes}\mathcal{O}_{\mathbb{C}^l}=\mathcal{O}_{\mathbb{C}^{k+l}}$) and denote its kernel by $I$. Then $I/I^2\cong \Omega^1$.
It confuses me that these two construction give two wildly different answer.
For example I tried to prove $de^x=e^xdx$ in $I/I^2$ and have not been able to do so. One has to prove that in $\mathcal{O}_M\hat{\otimes}\mathcal{O}_M$ the following holds
$1\hat{\otimes}e^x - e^x\hat{\otimes} 1 \mod I^2 = e^x(1 \hat{\otimes}x - x \hat{\otimes}1) \mod I^2$.
Moreover, however this proof works, it needs to fail if one tries to adapt the proof for the ideal $J$. Has someone worked this out explicitly or any ideas on how to do this? It should probably be a rather neat calculation.
In general, the tensor product $\mathcal{O}_M\otimes_{\mathbb{C}}\mathcal{O}_M$ is quite weird in the analytic category as it does not describe an analytic space. All of this probably just exemplifies that the fiber product in the algebraic category is different to the analytic fiber product.
 A: So, the calculation is actually quite simple. As @Yai0Phah correctly mentioned the analytic tensor product differs from the tensor product over $\mathbb{C}$ in the fact that infinite sums for convergent power series commute with the tensor product.
Now, note that
$(1\otimes x - x \otimes 1)(1\otimes x^n - x^n\otimes 1)= 1\otimes x^{n+1} - x\otimes x^n - x^n\otimes x + x^{n+1}\otimes 1$
vanishes in $I/I^2$. Expanding $e^x$ as a power series and then commuting the tensor product shows that all one has to show is:
$(\frac{1}{(i-1)!}x^{i-1} \otimes 1)(1\otimes x - x \otimes 1) \mod I^2 = (1\otimes \frac{1}{i!}x^i - \frac{1}{i!}x^i \otimes 1) \mod I^2$
But this follows from the vanishing of the above product in the following way:
\begin{align*}(1\otimes \frac{1}{i!}x^i - \frac{1}{i!}x^i \otimes 1) &= x\otimes \frac{1}{i!}x^{i-1} + \frac{1}{i!}x^{i-1}\otimes x - 2\frac{1}{i!} x^i\otimes 1\\
&=(x\otimes 1 )( 1\otimes \frac{1}{i!}x^{i-1} + \frac{1}{i!}x^{i-2}\otimes x - 2\frac{1}{i!} x^{i-1}\otimes 1)\\
&=(x\otimes 1 )( x\otimes \frac{1}{i!}x^{i-2} + 2\frac{1}{i!}x^{i-2}\otimes x - 3\frac{1}{i!} x^{i-1}\otimes 1)\\
&=(x^2\otimes 1 )( 1\otimes \frac{1}{i!}x^{i-2} + 2\frac{1}{i!}x^{i-3}\otimes x - 3\frac{1}{i!} x^{i-2}\otimes 1)
\end{align*}
Doing this $(i-1)$-times leads to
\begin{align*}(x^{i-2}\otimes 1)(x\otimes \frac{1}{i!}x + (i-1)x\otimes \frac{1}{i!}x - i \frac{1}{i!}x \otimes 1)&=(x^{i-1}\otimes 1) \frac{1}{(i-1)!}(1\otimes x - x \otimes 1).
\end{align*}
Which is what needs to be shown. All the above equations are to be considered $\mod I^2$.
