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Assume $V=L$ and let $\kappa$ be a Mahlo cardinal. Let $L[G]$ be the generic extension obatined by Mitchell forcing to make $2^{\aleph\_0}=\aleph\_2=\kappa.$
It is known that in the extension there are no special $\aleph\_2$-Aronszajn trees but there are $\aleph\_2$-Aronszajn trees. 



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> **Question 1.** Is there any $\aleph\_2$-Souslin tree in $L[G]?$
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I assume the believe is that there are, but I don't know how to prove it. Surprisingly, if, instead of $\aleph\_2$, we consider a cardinal $\lambda^+ > \beth\_\omega,$ with $\lambda$ regular and apply the Mitchell forcing to get 
$2^\lambda=\lambda^{++}=\kappa,$ then the results of Assaf Rinot show that there are $\lambda^{++}$-souslin trees in the extension.




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My second question is motivated by the work of Laver-Shelah. Assume $V=L$ and $\kappa$ is weakly compact. In order to produce a model of $CH+$there are no 
 $\aleph\_2$-Souslin trees, Laver and Shelah, first force with Levy collapse $Col(\aleph\_1, < \kappa),$ and over it do an iteration to kill all possible $\aleph\_2$-Souslin trees.



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> **Question 2.** Are there any $\aleph\_2$-Souslin trees just after doing the Levy collapse $Col(\aleph\_1, < \kappa)$?
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Two questions about higher Souslin trees
**About question 1:** If $\kappa$ is not weakly compact in $L$, then there is an $\aleph\_2$-Suslin tree in $L[G]$. 


In $L$ there is a $\kappa$-Suslin tree, $T$, and since Mitchell's forcing is $\kappa$-Knaster, it cannot add an antichain of cardinality $\kappa$ to $T$: If $\dot{\mathcal{A}}$ is a name for unbounded antichain in $T$, we can pick for every $\alpha < \kappa$, a condition $p\_\alpha$ in Mitcell's forcing, such that $p\_\alpha$ forces that the $\alpha$-th member of $\dot{\mathcal{A}}$ is some specific $t\_\alpha\in T$. Let $I\subseteq \kappa$ be an unbounded subset such that for all $\alpha, \beta\in I$, $p\_\alpha$ is compatible with $p\_\beta$ then $\{t\_\alpha \mid \alpha \in I\}$ form an antichain in $L$.




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**About Question 2:**
As it is stated in the comments by Assaf, the answer to question 2 is also affirmative.


