Jtdcftul

When we model an operational amplifier, we say that the output voltage $V_{out}$ is proportional to the difference between the inputs $V_pm$, $V_{out}=A(V_+-V_-)$. I've read that this voltage is always with respect to ground.

However, when we add a double supply to the op-amp, how is ground defined? Does the circuit take it to be $V_-$? Or maybe it is $(V_+-V_-)/2$?

"Ground" in this context is really a misnomer. It's only a reference. For example, take this circuit that you probably know:

^{simulate this circuit – Schematic created using CircuitLab}

and connect the "ground" to another control instead:

^{simulate this circuit}

You can think of it as being the exact same circuit with the exact same behavior, provided that the reference control stays constant. It simply centers on this new reference, and there's now no need for an explicit ground.

The actual rules for opamps are:

1. The output goes as far as it needs to, to keep the two inputs equal.


2. If the "+" input is higher than the "-" input, it goes up; if lower,
   it goes down.


3. If the output hits one or the other supply rail, it stops there.
   
   (actually a volt or two short unless it's explicitly a rail-to-rail design)

You'll notice that there's no mention of "ground" in any of that. Only comparing the two inputs to each other and the output to the supply rails. That's it!

Look at how input offset voltage is defined for that particular op-amp. Remember that's the input differential voltage (at a particular common mode voltage) that is required to drive the output to exactly zero.

Use the same ground reference.

In almost all cases, the maximum offset voltage can drive the output quite a bit more than between the supply rails so it doesn't much matter.

Your question is a good one. Since the differential gain of an opamp is typically very large 100 dB and up, it is often not a detail that is considered.

Take for example the following simulation of a highly idealized opamp, where each opamp has a different output reference potential:

The output is nearly identical for the three cases, and for basic design, the output reference potential can be ignored.

In cases where you would use the transfer function of the opamp $A(s)$ for stability and frequency analysis its also a mute point. Good design practices ensure all three reference potentials become one single AC ground due to decoupling/bypass capacitors.

It’s like building a house and deciding what is considered ground level (0 meters altitude).You could use a GPS or something to get the actual altitude, but who cares. Just pick a reference that makes sense for the design. Negative altitudes are obviously referring to the basement. Positive altitudes are obviously floors above ground. With your custom coordinate system, your blueprints become easy to read and the house is easier to build. So, if you instead arbitrarily define a certain elevation to be +10 meters, then you have just defined ground level as being 10 meters below the +10 meter mark. (Also, the meter is already defined.)

Electrical ground is just what you define to be 0V. That means you implicitly defined ground when you defined V+ and V-. If you have a 9V battery, you know that the positive terminal is 9V higher than the negative terminal, but you could call the positive terminal +6V and the negative terminal -3V, or whatever else you want. What is already defined is 1 volt and positive/negative. Lastly, the reference is chosen so that the circuit is easy to understand.