# Purity and six operations?

The six operations $f_!,f^!,f_*,f^*,\otimes,\mathcal Hom$ have the property that they preserve estimates on weights in one direction.

For $f_!,f^!,f_*,f^*$ I can see, that they don't preserve purity general. One can also give sufficient criteria for them to preserve purity, by ensuring them to be "equivalent" to their dual (smooth pullback, proper pushforward...).

What about the other operations $\otimes,\mathcal Hom$? What is an example of two pure sheaves, such that say $\cal F \otimes \cal G$ is not pure anymore?

Are there condtions that force $\otimes,\mathcal Hom$ to preserve purity? (By making them equivalent to their dual?)

I am most interested in the following situation:

Let $IC_x$ and $IC_y$ be two Bruhat constructible $IC$-sheaves on a partial flag variety. Then it is well known, that the space $Ext^{\bullet}(IC_x,IC_y)$ is pure. Does the sheaf version of this statement hold as well? Is the sheaf ${\cal Hom}(IC_x,IC_y)$ pure?

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Re your question of two pure sheaves whose tensor product is not pure: take an IC complex which is not pointwise pure (a simple example is provided by the $H^1$ of a degenerating family of elliptic curves over $\mathbb{C}$). Then tensor it with the skyscraper at a point where the IC is not pointwise pure. –  Geordie Williamson Jul 9 '13 at 17:22
Another (easier) example: take two smooth subvarieties which intersect in something reducible, then the tensor product of the constant sheaves on these two subvarieties will yield the constant sheaf on a reducible subvariety, which is not pure. –  Geordie Williamson Jul 9 '13 at 17:25
I guess you want the diagonal inclusion $X \hookrightarrow X \times X$ to be "non-characteristic" (terminology as in Kashiwara-Schapira) for the two definitions of tensor product to agree up to a shift/twist. –  Geordie Williamson Jul 10 '13 at 19:40
I don't think it will be true: take a rank two flag variety. Then the Schubert varieties corresponding to $st$ and $ts$ intersect in the union of $s$ and $t$ which is reducible. Hence my second (easier) example applies. –  Geordie Williamson Jul 11 '13 at 15:33
Just a minor addition to Geordie's comments: note that $M\otimes N = \Delta^*(M\boxtimes N)$ and $\mathcal{H}om(M,N) = \Delta^!(\mathbb{D}M\boxtimes N)$. –  Reladenine Vakalwe Jul 12 '13 at 15:28