Transformation invariance

A function f(x)f(\mathbf{x}) is said to be invariant under the transformation x↦φ(x)=(φ1(x),…,φn(x))\mathbf{x}\mapsto \varphi(\mathbf{x})=(\varphi_{1}(\mathbf{x}),\ldots,\varphi_{n}(\mathbf{x})) if applying the transformation leaves the function unchanged:

f(φ(x))=f(x)f(\varphi(\mathbf{x}))=f(\mathbf{x})

For example, consider the function f(x,y)=xyf(x,y)=xy and the transformation

φ:(x,y)↦φ(x,y;α)=(αx,yα)\varphi:(x,y)\mapsto \varphi(x,y;\alpha)=\left( \alpha x, \frac{y}{\alpha} \right)

If we apply it we get

f(φ(x,y;α))=αx1αy=xy=f(x,y)f(\varphi(x,y;\alpha))=\alpha x \frac{1}{\alpha}y=xy=f(x,y)

We get the original function again. When this happens we say that the function ff is invariant under the transformation (x→φx, y→φy)(x\to \varphi_{x},\ y\to \varphi_{y}). This denomination works specifically for that pair: the function ff is only invariant under that specific transformation. If you take a different function or a different transformation, the property is no longer necessarily true.

Another example:

f(x,y)=3y,{x→φx(x,y,α)=x+αy→φy(x,y,α)=yf(x,y)=3y,\quad \begin{cases} x\to \varphi_{x}(x,y,\alpha)=x+\alpha \\ y\to \varphi_{y}(x,y,\alpha)=y \end{cases}

ff is invariant under the transformation, but in a very specific way: it's invariant because the transformation only changes xx, but f(x,y)f(x,y) is actually independent of xx. As such, f(x,y)≡f(y)f(x,y)\equiv f(y) and so f(φy(x,y,α))=f(y)f(\varphi_{y}(x,y,\alpha))=f(y).