Given a ctm $M$, is there a forcing which does not add $\omega_1$sequences, and forces CH in $M[G]$?

Incidentally, questions of this type may be more appropriate at math.stackexchange, but I'm not entirely sure what the cutoff is. – Noah Schweber Aug 28 at 5:27

5My opinion is that any question about forcing is ontopic on MO. – Joel David Hamkins Aug 28 at 16:33
Unless I'm misunderstanding the question, the answer is no.
I assume that a forcing extension $M\subseteq M[G]$ "adds $\omega_1$sequences" if there is some map $f:\omega_1^M\rightarrow M$ which is in $M[G]$ but not $M$. If so, then we can argue as follows (with $M\models \neg$ CH, since the other possibility is uninteresting):
If $\omega_1^M$ is countable in $M[G]$, then we've added a new real, so a fortiori a new $\omega_1$sequence.
So suppose $\omega_1^M=\omega_1^{M[G]}$. If $M[G]\models$ CH, there is in $M[G]$ some surjection $f$ from $\omega_1^{M[G]}$ to $\mathbb{R}^M$; but since $\omega_1^M=\omega_1^{M[G]}$, the function $f$ is an $\omega_1$sequence in the sense of $M$, and since $M\models\neg$ CH it must be a new $\omega_1$sequence.
EDIT: As per Joel's comment below, note that forcing plays no role here: any time I can extend a model of set theory to one in which CH holds without adding $\omega_1$sequences (in the sense of the original model), the original model must also satisfy CH. The relevant notion of "extension" here is end extension: $M\subseteq_{end}N$ iff $M$ is a substructure of $N$ and for each $x\in M$ we have $\{y\in M: y\in^Mx\}=\{y\in N: y\in^Nx\}$ (that is, no objects in $M$ "get new elements" when we pass to $N$).
Exercise: show that forcing extensions are in fact end extensions!

2The OP didn't say that $M$ has $\neg$CH, although I suppose that would be the main case. What the argument shows is that CH is equivalent to the assertion that there is a forcing notion forcing CH without adding new $\omega_1$sequences. Indeed, forcing has nothing to do with it. For any model $M$, there is an extension $N$ with CH and no new $\omega_1$sequences if and only if CH is already true in M. – Joel David Hamkins Aug 28 at 16:36

2@JoelDavidHamkins To be fair, I think it's understating it to say that $M\models\neg$ CH is the "main case" since the other case is literally trivial. The point about extensions vs. forcing extensions is good, and I've added it to my answer. – Noah Schweber Aug 28 at 16:54