A linear map in $2D$ whose components are Hodge-dual to each other is conformal

Question

The following is a nice little lemma that I stumbled upon, that I thought might be useful for other people too:

Let $(V,g)$ be a $2$-dimensional oriented inner product space, and let $T:V \to \mathbb{R}^2$ be a linear map defined by $T(v)=(\alpha_1(v),\alpha_2(v))$, where $\alpha_1 \in V^*$ is non-zero and $\alpha_2=-\star_{V^*}\alpha_1$. Then $T$ is an orientation-reversing conformal map.

Here $\star_{V^*}:V^* \to V^*$ is the Hodge-star operator associated with the metric and orientation on $V^*$ induced by those on $V$.

Edit: I guess that this claim can be 'upgraded' to an 'if and only if' statement, i.e. $f=(\alpha_1,\alpha_2)$ is conformal$+$ orientation-reversing if and only if $\alpha_2=-\star_{V^*}\alpha_1$. I would be interested to see a proof for that.

The question is really about how to prove that conformality implies $\alpha_2=\pm \star_{V^*}\alpha_1$. (the right sign than should be easy to verify).

Question: How to prove this lemma in an elegant way?

(I have one proof and I guess there are other shorter approaches out there).

The basic intuition is that in even-dimensional spaces, the Hodge-dual acting on forms of degree which is half the dimension, maps each form to a form which is orthogonal to it:

$$\langle \alpha, \star_{V^*} \alpha\rangle \text{Vol}_V= \alpha \wedge \star_{V^*} (\star_{V^*} \alpha)=\pm \alpha \wedge \alpha=0.$$

Thus,$\langle \alpha_2,\alpha_1\rangle=-\langle \star_{V^*} \alpha_1,\alpha_1\rangle=0$. The fact that $\alpha_1 \perp \alpha_2$ is essentially equivalent to $T$ being represented (w.r.t orthogonal bases) via a matrix whose rows are orthogonal-i.e. a rescaling of an orthogonal matrix.

(However, my proof does not exactly follow this line of reasoning.)

Answer 1

Proof:

$T(v)=(\alpha_1(v),\alpha_2(v))$ implies that

$$| T(v)|^2=| \alpha_1(v)|^2+| \alpha_2(v)|^2.$$ Since the Hodge operator commutes with duals*, we have $\alpha_2(v)=-\star_{V^*}\alpha_1(v)=\alpha_1(\star_{V}v)$. Thus $$| T(v)|^2=| \alpha_1(v)|^2+| \alpha_1(\star_{V} v)|^2=| \alpha_1|^2|v|^2,$$

so $T$ is conformal. (In the last equality we have used the fact that when $|w|=1$ $\{w,\star_{V} w\}$ form an orthonormal basis for $V$).

To see that $T$ is orientation-reversing, note that if $|v|=1$, than

$$ T(v) \wedge T(\star_{V} v) = (\alpha_1(v),-\star_{V^*} \alpha_1(v)) \wedge (\alpha_1( \star_{V} v),-\star_{V^*} \alpha_1(\star_{V} v))=$$

$$ (\alpha_1(v), \alpha_1(\star_{V} v)) \wedge (\alpha_1( \star_{V} v), \alpha_1(\star_{V}\star_{V} v))=$$

$$(\alpha_1(v), \alpha_1(\star_{V} v)) \wedge (\alpha_1( \star_{V} v), -\alpha_1(v))=(\alpha_1(v)e_1+ \alpha_1(\star v)e_2) \wedge (\alpha_1( \star v)e_1-\alpha_1(v)e_2)=$$

$$ -(| \alpha_1(v)|^2+| \alpha_1(\star_{V} v)|^2)e_1 \wedge e_2=-| \alpha_1|^2 e_1 \wedge e_2,$$

so $$ T \wedge T(v \wedge \star_{V} v)=-| \alpha_1|^2 (e_1 \wedge e_2),$$ i.e. $\det T=-| \alpha_1|^2 <0$.

(Here we used the fact that $(v,\star_{V} v)$ form a positive orthonormal basis for $V$).

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*Actually, we do not really need to use the fact that the Hodge operator commutes with duals:

We can instead think of every $v \in V$ as an element in $\tilde v \in V^{**}$. Thus,

$$| T(v)|^2=| \alpha_1(v)|^2+| \star_{V^*}\alpha_1(v)|^2=| \tilde v(\alpha_1)|^2+| \tilde v(\star\alpha_1)|^2=|\alpha_1|^2|\tilde v|^2,$$

where in the last equality we used the fact that when $|\alpha|=1$, $\{\alpha,\star \alpha\}$ form an orthonormal basis for $V^*$.

Since $|\tilde v|=|v|$, we obtain $| T(v)|=|\alpha_1||v|$ as required.

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Edit:

Here is a proof for the other direction. Suppose that $T=(\alpha_1,\alpha_2)$ is conformal. Then, after normalizing, we can assume that

$|v|^2=| T(v)|^2=| \alpha_1(v)|^2+| \alpha_2(v)|^2.$ Representing $\alpha_1,\alpha_2$ as vectors $v_1,v_2$ via the musical isomorphism, we obtain

$$|v|^2=\langle v,v_1 \rangle^2+\langle v,v_2 \rangle^2.$$

Plugging in $v=v_1$ we deduce that $v_1 \perp v_2$. Plugging in $v=v_1$ we get $|v_1|^2=\langle v_1,v_1 \rangle^2=|v_1|^4$ so $|v_1|=1$, and similarly for $v_2$.

(The same proof shows that any set of vectors satisfying Parseval's identity is orthonormal). So, we showed $v_1,v_2$ is an orthonormal basis, hence so is $\alpha_1,\alpha_2$, which implies $\alpha_1=\pm \star \alpha_2$.