# Non-linear path between symplectic forms in $\mathbb{R}^4$

Give an example of a pair of symplectic forms $$\omega_0,\omega_1$$ in $$\mathbb{R}^4$$, which:

$$1)$$ induce the same orientation (i.e., the volume forms $$\omega_0\wedge\omega_0$$ and $$\omega_1\wedge\omega_1$$ provide the same orientation)

$$2)$$ have some degenerate convex combination (i.e., $$\omega_t:=(1-t)\omega_0+t\omega_1$$ is degenerate for some $$t\in[0,1]$$)

$$3)$$ admit a smooth $$1$$-parameter family of symplectic forms joining them (i.e., symplectic forms $$\eta_t$$ varying smoothly on $$t$$ with $$\eta_0=\omega_0$$ and $$\eta_1=\omega_1$$)

Consider the forms:

$$\omega_0:=dx\wedge dy+dz\wedge dw+dx\wedge dz$$ $$\omega_1:=dx\wedge dy+dz\wedge dw+4 dy\wedge dw$$

They define the same orientation and are both symplectic because:

$$\omega_0\wedge\omega_0=\omega_1\wedge\omega_1=2dx\wedge dy\wedge dz\wedge dw$$

Furthermore, we can check that $$\omega_t\wedge\omega_t=2(1-2t)^2dx\wedge dy\wedge dz\wedge dw$$, so $$\omega_t$$ is degenerate $$\Leftrightarrow t=1/2$$.

Geometrically, I have a strong feeling that we can find $$\{\eta_t\}_t$$ by taking the segment between $$\omega_0$$ and $$\omega_1$$ and making a slight deviation around the point $$\omega_{1/2}$$.

How could I do this formally?

The specific factors $$1$$ in $$1. dx \wedge dz$$ and $$4$$ in $$4. dy \wedge dw$$ were not important in order to make $$\omega_0$$ and $$\omega_1$$ symplectic forms with the same orientation. Change these two factors for $$t$$-dependent functions themselves, for instance

$$\omega_0(t) = dx \wedge dy + dz \wedge dw - 2(t-1/2) dx \wedge dz \, , \\ \omega_1(t) = dx \wedge dy + dz \wedge dw + 8(t-1/2) dy \wedge dw \, .$$

The forms $$\omega'_t = (1-t) \omega_0(t) + t \omega_1(t)$$ are such that $$\omega'_0 = \omega_0(0) = \omega_0$$ and $$\omega'_1 = \omega_1(1) = \omega_1$$, and they are all nondegenerate for $$t \in [0,1]$$ since $$\omega'_t \wedge \omega'_t = 2(1 + \underset{\ge \, 0 \mbox{ for } t \in [0,1]}{\underbrace{16t(1-t)(t-1/2)^2}}) \, dx \wedge dy \wedge dz \wedge dw \; .$$

(Dividing $$\omega'_t$$ by $$\sqrt{1 + 16t(1-t)(t-1/2)^2}$$ then yields a path $$\omega''_t$$ with constant associated volume form, if that matters to you.)

(i) Some intuition for the above solution goes as follow. Observe that the paths $$\omega_0(t)$$ ($$0 \le t \le 1/2$$) and $$\omega_1(t)$$ ($$1/2 \le t \le 1$$) are affine paths of symplectic forms respectively connecting $$\omega_0$$ and $$\omega_1$$ to $$\omega_{std} = dx \wedge dy + dz \wedge dw$$. Hence their concatenation $$\omega''_t$$ ($$0 \le t \le 1$$) is a piecewise affine path of symplectic forms between $$\omega_0$$ and $$\omega_1$$. Since $$\omega''_{1/2} = \omega_{std}$$ is symplectic, all 2-forms sufficiently closed to it are symplectic: it is thus possible to smoothen $$\omega''_t$$ near time $$t = 1/2$$ to obtain a smooth path $$\omega'''_t$$ of symplectic forms between $$\omega_0$$ and $$\omega_1$$. In fact, since the space of symplectic forms is locally convex, one way to realize the smoothing is -- roughly speaking -- by interpolating affinely between $$\omega_0(1/2 - \delta)$$ and $$\omega_1(1/2 + \delta)$$ for small $$\delta$$. This general intuition (about affinely interpolating between affine paths) is behind considering the above ansatz $$\omega'_t$$, which indeed turned out to be a solution.

(ii) The original post raised the question whether a slight perturbation of the affine path $$\omega_t = (1-t)\omega_0 + t \omega_1$$ near the degenerate form $$\omega_{1/2}$$ could yield a smooth path $$\omega'_t$$ by symplectic forms? As I'll explain below, this is indeed the case for the symplectic forms $$\omega_0$$ and $$\omega_1$$ from the question, but this is not a general fact.

Any (linear) 2-form $$\omega$$ on $$\mathbb{R}^4$$ is uniquely expressible in the form: $$\omega = \alpha \, dx \wedge dy + \beta \, dx \wedge dz + \gamma \, dx \wedge dw + \delta \, dy \wedge dz + \epsilon \, dy \wedge dw + \phi \, dz \wedge dw \, .$$ In this way, the space of (linear) 2-forms is identified with $$\mathbb{R}^6$$, via $$\omega \leftrightarrow (\alpha, \beta, \gamma, \delta, \epsilon, \phi)$$. One computes that $$\omega \wedge \omega = F \, dx \wedge dy \wedge dz \wedge dw$$ where $$F = \alpha \phi - \beta \epsilon + \gamma \delta$$. The space of degenerate 2-forms thus correspond to the variety $$[F = 0]$$.

Since $$F$$ is a homogenous degree-2 polynomial on $$\mathbb{R}^6$$, $$[F=0]$$ is a conic variety, i.e. invariant under dilation $$\omega \rightarrow \lambda \omega$$. The gradient is $$DF = (\phi, - \epsilon, \delta, \gamma, - \beta, \alpha)$$. Observe that $$DF$$ is normal to $$[F=0]$$ and $$DF \neq 0$$ away from the origin, hence $$[F = 0] \setminus \{0\}$$ is codimension-one cooriented open smooth manifold. Moreover, near any nonzero point $$p \in [F = 0]$$, the hypersurface $$[F=0]$$ separates a component of $$[F > 0]$$ from a component of $$[F < 0]$$.

Coming back to the affine path $$\omega_t = (1-t)\omega_0 + t \omega_1$$ of the original post, this path intersects $$[F = 0]$$ in only one point. Hence whether this path can be perturbed into a smooth path disjoint from $$[F=0]$$ is equivalent to asking whether $$\omega_0$$ and $$\omega_1$$ belong to the same connected component of $$\mathbb{R}^6 \setminus [F=0]$$. The above solution shows that this is indeed the case. Alternatively, one could merely note that $$\omega_t \in [F > 0]$$ for all $$t \in [0,1] \setminus \{ 1/2 \}$$: in other words, the affine path $$\omega_t$$ is tangent to $$[F=0]$$ at time $$t = 1/2$$ but does not cross this hypersurface, hence it can be perturbed to avoid the hypersurface altogether.

(iii) As a matter of fact, we have the following characterization:

Two (linear) symplectic forms on $$\mathbb{R}^4$$ are connected by a smooth path of (linear) symplectic forms if and only if they induce the same orientation on $$\mathbb{R}^4$$ i.e. have the same sign for $$F$$.

It suffices to prove that the open set $$[F \neq 0]$$ has precisely two connected components -- equivalently, that the intersection of $$[F \neq 0]$$ with the unit sphere $$S^5 \subset \mathbb{R}^6$$ has precisely two connected components. Since $$F|_{S^5}$$ attains a local minimum (resp. a local maximum) on each connected component of $$[F|_{S^5} < 0]$$ (resp. $$[F|_{S^5} > 0]$$), and since those local extrema are critical points of $$F|_{S^5}$$, it suffices to prove that the critical points of $$F|_{S^5}$$ with $$F < 0$$ (resp. $$F > 0$$) are connected to one another by a path in $$[F \neq 0]$$.

Note that $$p \in S^5$$ is a critical point of $$F|_{S^5}$$ iff $$DF(p)$$ and $$p$$ are colinear in $$\mathbb{R}^6$$. Since $$F(p) = \frac{1}{2} p^T Hp = \frac{1}{2} p^T \, DF(p)$$ with $$H = \mathrm{antidiag}(1, -1, 1, 1, -1, 1)$$, the critical points are exactly the eigenvectors of $$H$$. The eigenvalues of $$H$$ are $$\pm 1$$: the $$(+1)$$-eigenspace is generated by the vectors $$p_1 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{array} \right) \, , \, p_2 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ -1 \\ 0 \end{array} \right) \, , \, p_3 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 0 \\ 0 \\ 1 \\ 1 \\ 0 \\ 0 \end{array} \right)$$ while the $$(-1)$$-eigenspace is generated by the vectors $$p_4 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ -1 \end{array} \right) \, , \, p_5 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ 1 \\ 0 \end{array} \right) \, , \, p_6 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 0 \\ 0 \\ 1 \\ -1 \\ 0 \\ 0 \end{array} \right) \, .$$

Hence $$F|_{S^5}$$ has two whole 2-sphere worth of critical points: the unit 2-sphere $$S^+$$ spanned by $$p_1, p_2, p_3$$ and the unit 2-sphere $$S^-$$ spanned by $$p_4, p_5, p_6$$. Of course, $$F|_{S^{\pm}} = \pm 1/2$$. These two 2-spheres are both connected and they cover the set of critical points of $$F_{S^5}$$. QED

• Here we got a pair of 2-forms which have the non-degenerate convex sum, which defines a path. How can we now proceed to the pair that has a degenerate convex sum? Oct 10, 2023 at 17:38
• There is a path (non-smooth), just retracting both forms to dxdy + dzdw , but may be there is an elegant formula that I didn't see. Oct 10, 2023 at 17:58