I think the empty language is NP but I'm not sure if it is NP-Complete


You are correct that $\emptyset\in NP$, since we already know that $\emptyset$ can be decided in constant time (a TM that immediately rejects), and $DTIME(O(1))\subseteq P\subseteq NP$.

But $\emptyset$ is not NP-complete, regardless of whether $P=NP$. Indeed, by definition, a NP-complete language $A$ is such that every language $B \in NP$ admits a Karp reduction to $A$, i.e., a function $f$ that is computable in polynomial time and such that $x \in B \iff f(x) \in A$.

If $A= \emptyset$ and $B \neq \emptyset$ then no such function exists, since there is no way to satisfy $x \in B \implies f(x) \in A$.

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    $\begingroup$ +1 (of course if $P=NP$ then $\emptyset$ is $NP$-complete with respect to polytime Turing reductions, but that's a different thing). $\endgroup$ – Noah Schweber Apr 23 at 18:29

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