Why do I need static_cast and std::remove_reference in std::move if I pass an rvalue?

I'm trying to understand more move semantic. Now this is a possible implementation of std::move:

template <typename T>
typename std::remove_reference<T>::type&& move_(T&& obj)
{
    return static_cast<typename std::remove_reference<T>::type&&>(obj);
}

• But I've edited it a bit to understand its workings:

    template <typename T>
    typename std::remove_reference<T>::type&& move_(T&& obj)
    {
        if(std::is_lvalue_reference<decltype(obj)>::value)
            return static_cast<typename std::remove_reference<T>::type&&>(obj); 
        return obj; // dones't work?
        // return (string&&)obj; // ok works
    }
  
    auto s = move_(std::string("Hi there!")); // passing an rvalue cause instantiating: `string&&(string&&){}`. 
  

But why I can't directly return obj which is of revalue-reference type after instantiation?