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?


1 Answer 1


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.)

Additional comments:

(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

  • $\begingroup$ 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? $\endgroup$ Oct 10, 2023 at 17:38
  • $\begingroup$ 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. $\endgroup$ Oct 10, 2023 at 17:58
  • $\begingroup$ @LadaDudnikova Perhaps the comments I added above will help answer your questions? $\endgroup$ Oct 13, 2023 at 19:52
  • $\begingroup$ Very exciting, thanks. $\endgroup$ Oct 14, 2023 at 21:49

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