Consequently, the maps I, G and T ω are projective automorphisms of the cubic X 3 J . Let Conf(J) be the conformal group of J, that is the subgroup of P GL(Z 2 (J)) generated by I and the maps G g and T ω for all ω ∈ J and all g ∈ Str(J). From Lemma 4.6, it follows that Conf(J) is a subgroup of the group of projective automorphisms of X 3 J .Proposition 4.7. The group of projective automorphisms of X 3 J acts transitively on 3uple's of general points of X 3 J . In fact, if x 1 , x 2 , x 3 ∈ J are sufficiently general, it exists µ ∈ Conf(J) such that µ(ν 3 (x 1 )) = 0 J , µ(ν 3 (x 2 )) = 1 J and µ(ν 3 (x 3 )) = ∞ J .Proof. First let G = Str(J) · e be the orbit in J of the unit e under the linear action of the structural group. By [61, Theorem 6.5], one knows that G is a Zariski-open subset of J.Assume that x, y, z ∈ J are such that (1) y 1 = y − x and z 1 = z − x are invertible; (2) z 2 = z −1 1 − y −1 1 ∈ G , i.e z 2 = g(e) for a certain g ∈ Str(J). Then setWe leave to the reader to verify that µ(ν 3 (x)) = ∞ J , µ(ν 3 (y)) = 0 J and µ(ν 3 (z)) = 1 J . Since ν 3 (J) is dense in X 3 J , the conclusion follows.Let x 1 , x 2 , x 3 ∈ J and let µ ∈ Conf(J) be as in the statement of Proposition 4.7. Since µ ∈ P GL(Z 2 (J)), the curve µ −1 (C J ) is a rational normal curve of degree 3 passing through the points ν 3 (x i ) for i = 1, 2, 3. Since µ is a projective automorphism of X 3 J (by Lemma 4.6 above), this twisted cubic curve is also contained in X 3 J . Thus we have proved the following result.Corollary 4.8. The twisted cubic X 3 J associated to a Jordan algebra J of rank three is 3-covered by rational normal curves of degree 3.Let k be the dimension of a Jordan algebra J of rank 3. Then X 3 J is a non-degenerate algebraic subvariety of the projective space PZ 2 (J), whose dimension is 2k + 1. Since π(k, 3, 3) = 2k + 2, one obtains that X 3 J ⊂ PZ 2 (J) = P 2k+1 is an example of X(k, 3, 3). Thus the cubics X 3 J associated to Jordan algebras of rank 3 are examples of varieties of