Because you specifically ask for common-mode signals in differential amplifiers:
It is the purpose of a diff. amplifier to amplify ONLY the difference between two input signals with diff.gain Ad. However, this is not completely and fully possible (not ideal) because no circuit behaves ideally. Hence, the common-mode gain Acm is - more or less - a "quality figure" of a diff. amplifier. Low common mode gain gives a high quality figure - in particular we define the "common mode rejection ratio" CMRR=Ad/Acm (in dB).
In this context, it is important to know that each diff. signal (V1-V2) at the input - even if one input is zero - contains also a common-mode signal Vcm because both signals V1 and V2 can be split into two parts:
V1=Vcm+Vdd and V2=Vcm-Vdd
Vdd=(V1-V2)/2 and Vcm=(V1+V2)/2
Now - the diff amplifier should have a large diff gain Ad and a low common-mode gain Acm. For the classical diff. amplifier (long-tailed pair with two transistors) a very large CMMR (low gain Acm) can be achieved using a BJT current source in the common emitter leg (large fedback for common-mode signals Vcm).
EDIT/UPDATE: Explanation - the voltage component Vdd is called "push-pull signal" because it has different signs within V1 and V2. That means: The common-mode voltage Vcm can be seen as a arithmetic mean between V1 and V2, and Vdd is the component that must be added (subtracted) for gettuing V1 or V2, respectively.