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This is a sort of continuation of this question.

In synthetic differential geometry (SDG), we have $D\subset R$ comprised of the second order nilpotents. The Kock-Lawvere axiom (KL axiom) implies that a function $D\times D\to R^n$ is of the form $a_0+a_1d_1+a_2d_2+a_3d_1d_2$. This is like a 2-jet without square terms $f(a)+\partial_xf |_ad_1+\partial_yf|_ad_2+\partial_{xy}f|_ad_1d_2$.

In SDG, the infinitesimal rectangle $D\times D$ represents the second tangent bundle. In light of the KL-axioms I expect the classical second tangent bundle $\mathrm T^2X=\mathrm {TT}X$ of a $C^\infty$ manifold admits the following kinematic description: elements are equivalence classes of germs of $C^\infty$ maps $I^2\to X$ where $I$ is an interval about zero, and we identify such germs if upon composing with any germ in $C_{X,x}^\infty$ the partials and mixed partials coincide. Let us call such things "microsquares". They formalize the "2-jets without square terms" above.

If correct, this kinematic description is very geometric. For instance, it allows to define the flip on $\mathrm T^2X$ by flipping the $x,y$ coordinates of $I^2$. The two maps $\mathrm T^2X\rightrightarrows \mathrm TX$ given by $\mathrm T\pi_X,\pi_{\mathrm TX}$ are respectively given by restricting a microsquare to the $x$-axis and the $y$-axis. These fiber $\mathrm T^2X$ in two different ways: the fiber of $\mathrm T\pi_X$ over a kinematic tangent $\dot \gamma$ consists microsquare which restricts to $\gamma$ on the $x$-axis, and analogously for $\pi_{\mathrm TX}$.

The vertical lift applied to the tangent bundle gives a bundle isomorphism $\mathrm T(\mathrm TX/X)\cong \mathrm TX\times_X\mathrm TX$ over $\mathrm TX$, where the LHS is the vertical bundle of the tangent bundle, i.e the kernel of $\mathrm T\pi_X$. For all vector bundles this acts by taking a kinematic tangent (to a fiber of the bundle) to its derivative (which is a vector in the fiber).

Question 1. How to geometrically interpret the vertical lift for a "vertical microsquare"? A microsquare lies in the vertical bundle if its restriction to the $x$-axis is "constant", i.e the derivative of the restriction is zero. This is like saying the associated "2-jet without square terms" has $\partial_xf|_a=0$. What is the vertical lift doing with a microsquare that only makes sense if its restriction to the $x$-axis is zero?

My question is motivated by another one about a seeming discrepancy between SDG and the classical $C^\infty$ world:

  • In the $C^\infty$ world, the vertical lift $ \mathrm T(\mathrm TX/X)\cong \mathrm TX\times_X\mathrm TX$ is defined on any vertical microsquare. There is no further requirement for also being in the kernel of $\pi _{\mathrm TX}$ (restriction of a microsquare to its $y$-axis), and I see no reason for these kernels to coincide.

  • In SDG, the Wraith axiom says that a function $D\times D\to R^n$ which is constant on the axes uniquely factors through the multiplication map $D\times D\to D$. This factorization takes such a function to a tangent vector, and this is the analog of the vertical lift. The $C^\infty$ version of being constant on the axes is having the $\partial_x,\partial_y$ coefficients of the '2-jet without square terms' vanish $\partial_xf|_a=0=\partial_yf|_a$. The remaining mixed partial term indeed factors through the multiplication map because that's how Taylor series are. The point is that the Wraith axiom asks for both partials to vanish, as opposed to the vertical bundle which involves only vanishing $\partial_x$.

Question 2. What is going on here, geometrically? Why does SDG want both partials to vanish while the $C^\infty$ world only cares about one of the partials?

Lastly and perhaps most fundamentally: I don't understand the geometric meaning of a microsquare. I understand 2-jets since we retain the information of the Hessian, but retaining only the mixed partials - I don't get it.

Question 3. What is the geometric content of a microsquare / an element in the second tangent bundle?

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  • $\begingroup$ Perhaps related: Kochan-Severa's "differential gorms" arxiv.org/abs/math/0307303 (although it's a very different perspective) $\endgroup$ Nov 9, 2020 at 22:43
  • $\begingroup$ Dear @AlexArvanitakis, thanks for the reference. I feel my question is extremely elementary so I am hesitant to dive into that paper, at least for now. $\endgroup$
    – Arrow
    Nov 9, 2020 at 22:51

1 Answer 1

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I will try to address your questions, and then point to some general cartegorical phenomena that are at play here.

Answer 1/2: In the category of smooth manifolds, or a proper model of synthetic differential geometry where your base number line has negatives, the two axioms are equivalent for vector bundles. The Wraith axiom may seem stronger, but suppose you have $x: TE$ that it is over the kernel of $T\pi$ and so it splits as $(x_1,x_2):E \times_M E$ - then if it were over the kernel of both $T\pi, p_E$ you would have $x_2 = 0$, giving the Wraith axiom. Showing the other axiom holds using the Wraith axiom requires subtraction (this is why the Cockett and Cruttwell used this version of the universality of the vertical lift in their definition of tangent categories). As far as I can tell, one of the reasons the Wraith axiom was introduced in synthetic differential geometry was so that the Lie bracket on vector fields can be constructed, and given a connected the Dombrowski splitting theorem would hold $T^2M \cong T(M) \times_M T(M) \times T(M)$.

Answer 3: I think a good way to think about these things is based on the Weil functors approach, which can be found in Natural Operations in Differential Geometry. The first tangent bundle corresponds to an action by $R[x]/x^2$, the second by $R[x,y]/(x^2,y^2)$, and the bundle of 2-jets is represented by $R[x]/x^3$ which can be seen as the equalizer of endomorphisms on $R[x,y]/(x^2,y^2)$ (the identity $id$ and the flip $c$ that flips the variables $(x,y) \mapsto (y,x)$).

Generalities on the vertical lift: The universality of the vertical lift is interesting - I'm not sure I would call it a geometric condition, in my experience it seems more algebraic. Kirill MacKenzie showed that a lot of the properties of the vertical lift on the tangent bundle, or a vector bundle, are satisfied by the core of a double vector bundle (the core is the subbundle of the apex $E$ the projects down to $0$ on each of the side-bundles $E^H, E^V$). In fact, there's a general universal property for triple vector bundles that gives you exactly the Jacobi identities.

Strictly speaking, you don't need the local triviality properties of vector bundles. For any commutative semiring $R$, you an define the limit sketch $RBun$ (an $R$-module bundle), and a double $R$-bundle is a model of the sketch $RBun \otimes RBun$. The first thing you can do is observe that for double $R$-bundles in any complete category, you can take the core of the double $R$-bundle; if you chose a commutative ring, then you can prove the core satisfies the stronger vertical lift axiom (it satisfies the Wraith axiom by definition).

You can also see that vertical connections pop up here in a somewhat surprising way. The vertical lift can be seen as a coreflection of $R$-bundles into the category of double $R$-bundles; vertical connections are sections of this coreflection. Once again, if you chose a commutative ring $R$, this is (probably) sufficient to use an analogue of the Dombrowski splitting theorem.

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  • $\begingroup$ Dear Ben, thanks for your answer! I'm not sure how you answer my questions though: for Q1, I am asking for a geometric description of what the vertical lift does to a microsquare. For Q2 I don't understand how the requirement of being vertical in the $C^\infty$ world is equivalent to having both partials vanish in SDG. For Q3 I am asking for a geometric interpretation of microsquares. $\endgroup$
    – Arrow
    Nov 12, 2020 at 11:10
  • $\begingroup$ Dear Ben, could you perhaps take a look at this question involving the vertical lift? I don't see how the secondary vector bundle structure is related to the vertical lift. $\endgroup$
    – Arrow
    May 6, 2021 at 9:10

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