Let $f:Y\to X$ be a morphism of smooth varieties (complex analytic or algebraic as you wish). We have $$ T^* Y \overset{f_d}{\longleftarrow} Y\times_X T^*X \overset{f_\pi}{\longrightarrow} T^* X $$ and, for any holonomic $\mathcal{D}_X$-module, we have the estimate $$ Ch(Lf^* \mathcal{M}) \subset f_d f_\pi^{-1} Ch(\mathcal{M} ). $$

We know this is an equality when $f$ is non-characteristic for $\mathcal{M}$ meaning that $$ (f_d)^{-1} ( T_Y^* Y) \cap (f_{\pi})^{-1} Ch(\mathcal{M}) \subset Y\times_X T^*_X X $$

**Question** Is this an equality when $f$ is the blow-up of $X$ along a smooth sub-variety $A$?

I have checked this in a few simple cases but I'm having trouble proving it in full generality.

Edit: Here's another closely related question.

**Question** If $Z = \{g=0\}$ is a smooth hypersurface of $X$. What is a necessary and sufficient condition for the proprer transform of $Z$ to be non-characteristic to $Ch(Lf^*\mathcal{M})$ in terms of $Z$, $A$ and $Ch(\mathcal{M})$?

A typical example would be $X= \mathbb{A}^2$, $A = \{0\}$, $Ch(\mathcal{M})$ the Lagrangian variety corresponding to the stratification of $\mathbb{A}^2$ by the axises $x=0$, $y=0$ (i.e. the union of the zero section of $T^*X$, the conormal bundles to the axises and the cotangent space at 0). Then the proprer transform of an hypersurface $Z$ will be non-characteristic iff $Z$ is clean with respect (i.e. non tangent) to the axis.

`T^*_Y`

, but thought`T_Y^*`

was fine. – Ben Webster♦ Feb 15 '11 at 1:06