Lambda function, strange behavior

Let's assume we have the following lambda declared in the global namespace:

auto Less = [](int a,int b) -> bool
{
    return a < b;
}

And the following code which make use of this lambda:

template<typename T>
struct foo
{
    foo(int v){}
    bool operator<(const foo<T>&) const
    {
        return T(1,2);
    }
};

int main()
{
    typedef foo<decltype(Less)> be_less;
    priority_queue<be_less> data;
}

As you can see i use Less as template parameter for the struct foo. With g++4.9.2 this code does not compile:

test1.cpp:13:21: error: no matching function for call to '<lambda(int, int)>::__lambda0(int, int)'
     return T(1,2);
                 ^
test1.cpp:13:21: note: candidates are:
test1.cpp:17:14: note: constexpr<lambda(int, int)>::<lambda>(const<lambda(int, int)>&)
 auto Less = [](int a,int b) -> bool
          ^
test1.cpp:17:14: note:   candidate expects 1 argument, 2 provided
test1.cpp:17:14: note: constexpr<lambda(int, int)>::<lambda>(<lambda(int, int)>&&)
test1.cpp:17:14: note:   candidate expects 1 argument, 2 provided

Bu i may fix this problem adding two small changes, first of all i change the lambda into this:

bool Less = [](int a,int b) -> bool
{
    return a < b;
}

As you can see i just replaced auto with bool, this modification alone still does not work:

test1.cpp:13:21: error: expression list treated as compound expression in functional cast [-fpermiss
ive]
     return T(1,2);

Unless i add -fpermissive, or i may change the operator< bool in this way:

bool operator<(const foo<T>&) const
{
    return T((1,2));
}

Note the double parentheses. Now the code compile and everything work.

My question is, which is the technical reason why auto Less does not work but bool Less work?

I believe i know why the double parentheses are required in the second operator<, this should be to avoid the compiler to interpret T(1,2) as a declaration instead of a call.

Thanks for your time