Use of decltype Gives warning Reference to Local Variable

I am trying to understand how decltype works in C++ and so going through examples. Just when i thought i understood what it means, i tried to create examples to confirm that and i got something that i cannot understand. The example is as follows:

#include <iostream>


template<typename T> 
auto func2(T a, T b) -> decltype(b < a ? a:b)
{ 
    //return a;//this gives warning of reference to local variable
    //return b;//this gives warning of reference to local variable
    
    return (b < a ? a:b);//this does not give warning of reference to local variable
}
int main()
{
   std::cout << "Hello World" << std::endl; 
   int b = func2(4,3);
   return 0;
}

As shown in the above code when i use the statement return a; or return b; i get a warning saying reference to local variable which is because we should never pass reference or pointers to local variable to outside their scope. But what i don't understand is that why are we not getting the same warning when i use return (b < a ? a:b);.

My thinking is that decltype(b < a ? a:b) should result in int&. And so when i write return (b < a ? a:b); a reference to local a or b should be returned depending on whether which one is greater and we should get the warning.

Check out the code here