Projective plane does not minimally immerse into $S^3$

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Can someone please give me a reference (or a proof) of the fact that the projective plane cannot be minimally immersed into the 3-sphere?

My reference is: Lawson, “Complete Minimal Surfaces in $\mathbb S^3$”, see Corollary 1.6. In this paper, among other things, Lawson explains the relation between the zeros of the Hopf holomorphic differential form on a minimal surface in $\mathbb S^3$ and the topology of the surface, which seems to bee the idea that leads to Corollary 1.6, my desired claim. However, I miss the logical step to prove the claim, maybe somebody can explain it to me. It is anyway clear to me that the projective plane cannot be embedded in $\mathbb S^3$ and that it can be immersed in $\mathbb S^3$.

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Assume by contradiction the projective plane can be minimally imeresed. Then we can lift this immersion to a minimal immersion $\psi: S^{2} \to S^3$, as $S^{2}$ is its orientable double cover.

Proposition 1.5 in your reference is: Let $\psi: \mathcal{R} \rightarrow S^{3}$ be a minimal immersion where $\mathcal{R} $ is compact and of genus $g .$ Then

  1. (F. Almgren) If $g=0$, the immersion is totally geodesic.
  2. If $g \geqq 1$, then $\sum_{p \in \mathscr{R}} d_{p}=4 g-4$.

Thus our immersion $\psi: S^{2} \to S^3$ of the double cover is totally geodesic. To reach a contradiction take a great circle on $S^2$ it will be sent to a geodesic in $S^3$ as $\psi(S^2)$ is totally geodesic. Do you see how to get the contradiction by thinking on the corresponding geodesic of the projective plane.