How to write Subqueries with In-Expressions in JPA 2.0, Criteria API?

Late resurrection.

Your query seems very similar to the one at page 259 of the book Pro JPA 2: Mastering the Java Persistence API, which in JPQL reads:

SELECT e 
FROM Employee e 
WHERE e IN (SELECT emp
              FROM Project p JOIN p.employees emp 
             WHERE p.name = :project)

Using EclipseLink + H2 database, I couldn't get neither the book's JPQL nor the respective criteria working. For this particular problem I have found that if you reference the id directly instead of letting the persistence provider figure it out everything works as expected:

SELECT e 
FROM Employee e 
WHERE e.id IN (SELECT emp.id
                 FROM Project p JOIN p.employees emp 
                WHERE p.name = :project)

Finally, in order to address your question, here is an equivalent strongly typed criteria query that works:

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Employee> c = cb.createQuery(Employee.class);
Root<Employee> emp = c.from(Employee.class);

Subquery<Integer> sq = c.subquery(Integer.class);
Root<Project> project = sq.from(Project.class);
Join<Project, Employee> sqEmp = project.join(Project_.employees);

sq.select(sqEmp.get(Employee_.id)).where(
        cb.equal(project.get(Project_.name), 
        cb.parameter(String.class, "project")));

c.select(emp).where(
        cb.in(emp.get(Employee_.id)).value(sq));

TypedQuery<Employee> q = em.createQuery(c);
q.setParameter("project", projectName); // projectName is a String
List<Employee> employees = q.getResultList();