Assume $V=L$.


Let $\alpha$ be the least ordinal such that there is a $\Diamond\_{\omega\_1}$-sequence in $L\_\alpha$.


It's obvious that $\omega\_1 < \alpha < \omega\_2$.


Do we have some better estimates of $\alpha$?


Also, what about $\Diamond^+$, squares and morasses?



Height of diamond
It depends on what diamond sequence you have.


$V=L$ proves there is a $\diamondsuit$-sequence of $L$-rank $\omega\_1+1$: recall the famous way of proving the existence of $\diamondsuit$-sequence. The definition goes as follows (for example, Theorem 13.21 of [Jech - Set theory (3rd edition)](https://doi.org/10.1007/3-540-44761-X)):



> 
> $(S\_\alpha,C\_\alpha)$ is the $<\_L$-least pair $S\_\alpha\subseteq\alpha$, $C\_\alpha\subseteq\alpha$ a club, and $S\_\alpha\cap\xi\neq S\_\xi$ for all $\xi\in C\_\alpha$. Otherwise $S\_\alpha=C\_\alpha=\alpha$.
> 
> 
> 


Observe that if $X\subseteq\alpha<\omega\_1$, then $(\alpha,X)\in L\_{\omega\_1}$. Furthermore, $<\_L$ is absolute between $L$ and $L\_{\omega\_1}$. Combining with that $L\_{\omega\_1}$ satisfies $\mathsf{ZFC}^-$, which implies $L\_{\omega\_1}$ is closed under recursively defined sequences over there, we can see that the above definition works over $L\_{\omega\_1}$. Hence the $\diamondsuit$-sequence defined as above is a member of $\operatorname{Def}(L\_{\omega\_1})=L\_{\omega\_1+1}$.


However, the $L$-rank of a $\diamondsuit$-sequence can be arbitrarily large for a silly reason. Observe that the $(\alpha+1)$-th component of $\diamondsuit$-sequence does not affect being $\diamondsuit$, and hence we may replace it arbitrarily.


Let $f:\omega\_1\to\omega\_1$ be the increasing enumeration of successor ordinals below $\omega\_1$. (Note that $f\in L\_{\omega\_1+1}$.) Recall the $\diamondsuit$-sequence $\langle A\_\alpha\mid \alpha<\omega\_1\rangle$ we previously defined (so of $L$-rank $\omega\_1+1$.) For any $B\subseteq \omega\_1$, define $A^B\_\alpha$ by
$$A^B\_{f(\alpha)}=\begin{cases} 0&\text{if }\alpha\in B \\ 1 & \text{if } \alpha\notin B\end{cases}$$
and $A^B\_\alpha=A\_\alpha$ for a limit $\alpha$. Then we can recover $B$ from $\langle A^B\_\alpha\mid\alpha<\omega\_1\rangle$ (possibly except for information about $0\in B$ and $1\in B$, but it does not matter.) Hence the $L$-rank of $\langle A^B\_\alpha\mid\alpha<\omega\_1\rangle$ is greater than or equal to that of $B$. (I suspect they are equal.)


*There is no reason to believe that the $L$-rank of a $\diamondsuit$-sequence is **equal** to a given ordinal less than $\omega\_2$. I think the question [A tree coding all of $L\_\alpha$ within $L\_\alpha\cap\mathcal{P}(\omega)$](https://math.stackexchange.com/q/4072513/53976) in MSE indicates, at least, the above construction does result in a $\diamondsuit$-sequence whose $L$-rank is equal to a given ordinal.*


I believe the same argument applies for $\diamondsuit^+$-sequences, but I have not checked in detail. I am not sure the same holds for $\square$-sequences or morasses.


