super constraint

If I were using the syntax that seems like it should work:

public class SomeClass<T>
{
    public SomeClass<SuperT> SomeMethod<SuperT>() where T : SuperT
    {  }
}

If I were writing the method in Java:

public class SomeClass<T>
{
    public <SuperT super T> SomeClass<SuperT> SomeMethod()
    {  }
}