Provide a condition on which, We have $f^*(T)=T$

Question

Definitions:

A function $f:V^k\to \mathbb R$ is called a $k$-tensor if it takes $k$ members of a vector space $V$, as the input, and is linear with respect to each of the input elements.
For example, $f(\alpha v_1,\dots,v_k)=\alpha f(v_1,\dots,v_k)$ and $f(v_1+v',\dots,v_k)=f(v_1,\dots,v_k)+f(v',\dots,v_k)$

$L^k(V)$ denotes the set of all $k$-tensors on the vector space $V$.

We know that $\{e^i \otimes e^j : i,j\in\{1,\dots,n\}\}$ is a basis of $L^2(V)$ such that $V$ is a $n$-dimensional vector space and $e^i(e_j)=\delta_{ij}$.

If $\phi:V \to W$ is a linear transformation, $\phi^*:L^1(W)\to L^1(V)$ such that $\forall f \in L^1(W)\quad \phi^*(f)=f \circ \phi$, Is called the pullback of $\phi$.

---

Question:

On $\mathbb R^2$, Assume that $T=e^1 \otimes e^2 - e^2\otimes e^1$ and $f:\mathbb R^2\to \mathbb R^2$ is a function such that $f(x,y)=(ax+by,cx+dy)$ . Can you provide a condition on which, We have $f^*(T)=T$ ?

My first thoughts (If $T$ is from $\mathbb R^2$ to $\mathbb R$):

$T \circ f = T \implies \forall (x,y)\in \mathbb R^2\quad T(x,y)=T(f(x,y))=T(ax+by,cx+dy)\implies \forall (x,y) \in \mathbb R^2\quad (e^1\otimes e^2-e^2\otimes e^1)(x,y)=(e^1\otimes e^2-e^2\otimes e^1)(ax+by,cx+dy)$
This is clearly wrong because $(e^1\otimes e^2-e^2\otimes e^1)$ is from $\mathbb R^2 \times \mathbb R^2$ to $\mathbb R$.

My second thoughts (If $T$ is from $\mathbb R^2 \times \mathbb R^2$ to $\mathbb R$):

$T \circ f=T \implies \forall (a,b,c,d) \in \mathbb R^4 \quad T(a,b,c,d)=T(f(a,b,c,d))$

Again this is wrong because $f$ is from $\mathbb R^2$ to $\mathbb R^2$, So $f(a,b,c,d)$ is not defined.

What am i missing??!

Answer 1

The effect of your linear map is $$f(e_1)=(a,\, c)=ae_1+ce_2, \quad f(e_2) = (b, \, d)=be_1+de_2,$$ so with simple computations involving bilinearity we have $$\begin{split}(f^*T)(e_1, \, e_1) & = T(ae_1+ce_2, \, ae_1+ce_2)=ac-ca=0, \\ (f^*T)(e_1, \, e_2) & = T(ae_1+ce_2, \, be_1+de_2)=ad-cb, \\ (f^*T)(e_2, \, e_1) &= T(be_1+de_2,\, ae_1+ce_2)=bc-da, \\ (f^*T)(e_2, \, e_2) & = T(be_1+de_2, \, be_1+de_2)=bd-db=0 \end{split}$$ On the other hand, we also have $$T(e_1, \, e_1)= 0, \quad T(e_1, \, e_2)= 1, \quad T(e_2, \, e_1)= -1, \quad T(e_2, \, e_2)= 0,$$ so $$f^*T=(ad-bc)T.$$ In particular, $f^*T=T$ if and only if $ad-bc=1$.