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?