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

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.)

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 $$V$$).

*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.

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 $$\alpha_1=\pm \star \alpha_2$$.