# What is the exterior derivative intuitively?

Actually I have several related questions, not worth opening different threads:

1. What is the exterior derivative intuitively? What is its geometric meaning? A possible answer I know is, that it is dual to the boundary operator of singular homology. However I would prefer a more direct interpretation.

2. What is a conceptually nice definition of the exterior derivative?

• Apr 11, 2010 at 19:11

Many years back I wrote something about an intuitive way to look at differential forms here. In particular, figure 4 illustrates Stokes' theorem in a way that generalises to higher dimensions. Note that these are just sketches for intuition, and I've found them useful for illustrating various fields arising in physics, but they're not anything rigorous. They're also, in some sense, dual to the diagrams in Misner, Thorne and Wheeler. (There are some errors in my document, but I lost the source code many years ago...) • Very nice indeed. I strongly recommend reading the linked file. Feb 19, 2011 at 22:48
• Using these sorts of pictures, how do you see that the derivative of $dx + y\;dz$ is $dy \wedge dz$? Can you even draw a picture for $dx + y\;dz$? These pictures seem to characterize differential forms as "foliations of varying density," but I'm pretty sure that's only true for forms that are locally a function times an exact form, and there are lots of forms—for example, $dx + y\;dz$—that don't look like this. Nov 6, 2014 at 1:52
• By the way, in case you haven't seen it already, Gabriel Weinreich's Geometrical Vectors introduces differential forms and the exterior derivative from the perspective you describe. Nov 6, 2014 at 1:57
• The visualizations you refer to in Misner, Thorne, and Wheeler were originated by J.A. Schouten, and first presented in Ricci-Calculus: An Introduction to Tensor Analysis and its Geometrical Applications amazon.com/… . William L. Burke has given more concise and modern presentations, e.g., in Applied Differential Geometry.
– user21349
Jan 18, 2016 at 16:11
• @Asvin I've made a new link Sep 5, 2017 at 13:52

I think that the best explanation is in Arnold's book "Mathematical methods of classical mechanics". Here it is: after fixing a chart on a manifold one can say that the value of $d\omega$ ($\omega$ is a n-form) on tangent vectors $(\xi_1, ...,\xi_{n+1})$ at point $x_0$ equals to the coefficient of the $(n+1)$-linear part of the function $F(\varepsilon)=\int_{\partial V(\varepsilon)} \omega$, where $V(\varepsilon)$ is a "curvilinear parallelepiped" with vertexes $x_0, x_0+\varepsilon \xi_1, ..., x_0+\varepsilon \xi_{n+1}$: $F(\varepsilon)=(d\omega)(x_0)(\xi_1, ...,\xi_{n+1})\varepsilon^{n+1}+o(\varepsilon^{n+1})$.

• This is a good explanation too!
– j.c.
Apr 11, 2010 at 20:29
• In my opinion this is really "the" interpretation. Note that it treats functions and forms on an equal footing: the definition of the derivative of a function is exactly given by the above. Proving that $d\omega$ is in fact a form from this definition is also quite enlightening. Pedagogically, it is also great. It basically says "$d\omega$ is what makes stokes theorem true on 'infinitesimal' parallelepipeds". Dec 20, 2013 at 22:13
• It is an extension of the definitions in vector calculus, such as $\mathrm{div}F(p)=\lim\frac1{\pi r^2}\int_{\partial D_r(p)}F\cdot n$. Nov 14, 2019 at 22:04

For 1-forms, you can get some intuition for exterior differentiation from how it shows up in Frobenius's theorem which states that a distribution D is integrable if and only if the ideal of differential forms that are annihilated by it is closed under exterior differentiation:

Let $\alpha$ be a 1-form on $M$. If $\alpha$ does not vanish, then ker $\alpha_x$ is a hyperplane in the tangent space to $M$ at $x$. Thus ker $\alpha$ is a hyperplane field in $TM$ (and is an example of a distribution). At every point in M, you should visualize a hyperplane passing through that point.

Frobenius's theorem gives conditions on whether this hyperplane field is integrable, that is, if one can fit the planes together to form a foliation by hypersurfaces in $M$. For a hyperplane field defined by a single 1-form one can fit the planes together if and only if $d\alpha$ mod $\alpha$ is zero. This is usually expressed by the vanishing of $\alpha\wedge d\alpha$.

(In the general case, where instead of $\alpha$ we have a set of linearly independent 1-forms $\{\alpha_j\}_{j=1}^r$, the ideal in the algebra of differential forms on $M$ generated by $\{\alpha_j\}_{j=1}^r$ must be closed under exterior differentiation; equivalently $d\alpha_j\wedge\alpha_1\wedge\cdots\wedge\alpha_r=0$ for all $j$).

Two simple examples:

(1) if $\alpha=df$ then the field of hyperplanes ker $\alpha$ is actually tangent to the hypersurfaces $f=$const (and of course $d\alpha=0$).

(2) If $\alpha = g df$ for some non-vanishing function $g$, e.g. $\alpha=ydx$ in the upper half plane of $\mathbb{R}^2$, then this is just as good, since ker $\alpha$ is still tangent to $f=$const. Note that $d\alpha=dg\wedge df=(dg/g)\wedge\alpha$, which vanishes mod $\alpha$ and thus $\alpha\wedge d\alpha=0$.

Hence $\alpha\wedge d\alpha$, or $d\alpha$ mod $\alpha$ roughly measures how far this hyperplane field defined by ker $\alpha$ is from being tangent to hypersurfaces.

(I got the ideas from Appendix B of Ivey and Landsberg's book Cartan for Beginners. Thanks to Marcos Cossarini and Ben McKay for pointing out in the comments that the original version of this was wrong!)

Here's an example of a hyperplane field which is not tangent to any hypersurfaces. $\alpha = dz-y dx$ on $\mathbb R^3$ and $\alpha\wedge d\alpha = dz\wedge dx \wedge dy$: • Great answer! I wish I could accept both, your answer and sigfpe's. These two answers also fit nicely together. Apr 11, 2010 at 20:28
• I wonder if there's a way to "see" the exterior derivative for k-forms with k>1 along these lines, see e.g. t3suji's comment here mathoverflow.net/questions/12266/…
– j.c.
Apr 11, 2010 at 20:34
• If M=R^2, the 1-forms w=dx and q=ydx determine the same (integrable) distribution of hyperplanes, but dq is not zero. Perhaps the proposition should be weaker: A 1-form w determines an integrable distribution iff a non vanishing real-valued function f exists such that d(fw)=0. (¿Is this true?) Jul 7, 2010 at 3:19
• Not quite right. Integrability of the kernel of $\alpha$ is measured by $\alpha \wedge d\alpha$, not by $d\alpha$. Feb 21, 2013 at 21:28
• After 5 and a half years, I've finally fixed this. Thanks!
– j.c.
Sep 23, 2015 at 13:19

The exterior derivative is the unique (sequence of) linear map $d: \mathcal{A}^p (M) \to \mathcal{A}^{p+1}$, such that the following axioms hold:

1. for a function $f$, $df$ is the total differential.
2. For any function $f$ and any differential form $a$, the Leibniz rule $d(fa)= df \wedge a + f da$ holds.
3. For any diffeomorphism $\phi: M \to N$, you have $\phi^{\ast} \circ d = d \circ \phi^{\ast}$.

I think that 3 is more natural or at least easier to motivate than the usual $dd=0$. But both properties are really equivalent.

Proof (of uniqueness): 2. implies locality, i.e. the value of $d a$ at a point $x \in M$ only depends on the value of $a$ in a neighborhood of $x$. This, together with the axiom 3, shows that it is enough to consider $M =\mathbb{R}^n$.

The group $\mathbb{R}^n$ acts by translations on $\mathbb{R}^n$. By axiom 3, for any translation-invariant form $a$ on $\mathbb{R}^n$, the form $da$ is again translation-invariant.

On the other hand, each nonzero $\lambda \in \mathbb{R}$ gives rise to the diffeomorphism $h_{\lambda}:x \mapsto \lambda x$ of $\mathbb{R}^n$. It is easy to check that it acts on translation-invariant $p$-forms by multiplication with $\lambda^p$. Thus for any translation-invariant $p$-form $a$, you get

$$\lambda^p d a = d (\lambda^p a) = d (h_{\lambda}^{\ast} a ) = h_{\lambda}^{\ast} d a = \lambda^{p+1} da,$$

which implies that any translation-invariant form is closed. Finally, note that any $p$-form on $\mathbb{R}^n$ can be written as a linear combination of translation-invariant form, with coefficients in $C^{\infty}(\mathbb{R}^n)$ (a basis for the translation-invariant forms is formed by the usual elements $dx_{i_1} \wedge \ldots \wedge x_{i_p}$).

From axioms 1 and 2, you now conclude that $d$ must be the exterior derivative that you knew before. This, of course, implies all the other properties of $d$.

• This is great! It's the first definition of exterior differentiation that ever really made sense to me. I think I'll be using this one from now on. Sep 14, 2012 at 21:05
• I teach this (among other answers) and I also like to assign as a challenge that, if we just ask for a map $d : \Omega^1 \to (\Omega^1)^{\otimes 2}$ obeying (1)-(3), then it is forced to land in the alternating tensors. (Presumably, a similar argument applies for any $p$, but $p=1$ seems hard enough for the students.) Sep 5, 2017 at 14:19
• In fact, property 3 already characterizes the exterior derivative uniquely (up to a scalar multiple), see "Natural operations on differential forms" by Palais. Jun 9, 2020 at 18:29

There is a following (it seems to me it is not well-known but interesting) approach to differential forms. I'll try to reproduce it here. In this approach the exterior derivative is a very simple operation.

What is a differential k-form on a manifold $$M$$? Consider a (k+1)-product $$V_{k+1}(M)=M\times...\times M$$. Denote by $$S_k(M)$$ the space of all smooth skew-symmetric (with respect to a product structure) real functions on $$V_{k+1}$$. Obviously any function from $$S_k(M)$$ equals to zero on the diagonal $$\Delta=$$ {$$(x,x,...,x)| x\in M$$}.

We define a subspace $$L_k(M) \subset S_k(M)$$ as follows: $$L_k(M)$$ consists of all elements of $$S_k(M)$$ of order smaller then $$k$$ along the $$\Delta$$. In other words, $$f\in L_k(M)$$ if and only if for any smooth path $$I(t)$$ starting on the diagonal (i.e. $$I(0)\in \Delta$$) holds $$f(I(t))=o(t^k)$$.

Then one can identify the space of all k-forms $$\Omega_k(M)$$ with a quotient $$S_k(M)/L_k(M)$$.

What is the exterior derivative? Consider the following operator $$\delta: S_k(M)\to S_{k+1}(M)$$, $$\delta f(x_1,...,x_{k+2}) =\sum (-1)^{i+1} f(x_1,..,\hat{x_i},...,x_{k+2})$$. One can check that $$\delta (L_k(M))\subset L_{k+1}(M)$$ and that the induced operator $$\Omega_k(M)=S_k(M)/L_k(M)\to S_{k+1}(M)/L_{k+1}(M)=\Omega_{k+1}(M)$$ coincides with the exterior derivative $$d$$.

I know that approach from B.L. Feigin's lectures on multidimensional calculus (in Russian here: http://ium.mccme.ru/f98/calcman.html).

• This is cool! Am I right in thinking that you can identify the set of k-vectors on M with the set of derivations on S_k, just as you can identify the set of 1-vectors on M with the set of derivations on S_1? Apr 12, 2010 at 2:46
• This seems to me analogous to the way algebraic geometers define $\Omega^{1}$ as $\mathcal{I}_\Delta/\mathcal{I}_\Delta^{2}$, where $\mathcal{I}_\Delta$ is the ideal sheaf of the diagonal. Apr 12, 2010 at 16:39
• At the end of your third paragraph, should $I(t) = \mathrm o(t^k)$ be $f(I(t)) = \mathrm o(t^k)$? Jan 19, 2016 at 15:19
• I'd really like to read more about this approach, but unfortunately I can't read Russian. Do you happen to know of a reference in English, Spanish, or (as a last resort) some other Romance language? The closest thing I've found is Anders Kock's writing on synthetic differential geometry (for example, Section I.18 of Synthetic Differential Geometry), but it carries a lot of very general baggage; I just want to learn about plain old smooth manifolds. (As a very last resort, could you point me to where in Feigin's lectures this stuff can be found?) Nov 17, 2017 at 21:34
• @Vectornaut: It is in Lecture 9. The $\delta$ is defined in Lecture 8. Nov 18, 2019 at 1:21

For 2: it is the unique extension of the total differential $d:C^\infty(M)\to\Omega^1(M)$ to a graded derivation of the algebra $\Omega^\bullet(M)$ of differential forms.

The map $d:C^\infty(M)\to\Omega^1(M)$ itself has a nice characterization as a universal derivation of the algebra $C^\infty(M)$ of functions satisfying certain rather reasonable conditions---this follows from Jaak Peetre's theorem.

• The definition of a graded derivation was originally just a natural generalisation of $d$, so this approach is almost circular, and I can't visualise it geometrically. Feb 21, 2013 at 21:33
• Well, my point is that you need only visualize the component in degree zero, as the rest is simply formalities. Feb 21, 2013 at 23:27
• @Mariano Suárez-Alvarez Do you mean that for some subalgebra $A$ of $C^{\infty}(M)$ and $\Omega^1(M)$ truncated to be $A-$module $(A,\Omega^1(M))$ is Kähler differential? Could you give some refrences to your statement? May 26, 2015 at 14:24
• @Mariano Suárez-Alvarez The sentence: "The map d:C∞(M)→Ω1(M) itself has a nice characterization as a universal derivation of the algebra C∞(M) of functions satisfying certain rather reasonable conditions" May 26, 2015 at 15:40
• @FallenApart, ah. No , I do not mean that $\Omega^1(M)$ is the module of Kähler differentials of $C^\infty(M)$ (mostly, because it isn't! :) ) The operator $d:C^\infty(M)\to\Omega^1(M)$ can be characterized in terms of its functorial properties. This is surely done in detail in the book Natural Operations in Differential Geometry by Kolar, Michor and Slovak. May 26, 2015 at 16:14

To start with 0-forms, $df$ codes how $f$ varies. In fact, it does this in a way that is, IMO, more natural than partial derivatives.

For example, if I want to know how $z = x^2 y$ varies with $x$ — actually that's a lie, I want to know how $z$ varies with $x$ as $y$ is held constant — then I compute an exterior derivative, setting $dy=0$:

$$dz = 2xy \, dx + x^2 \, dy \equiv 2xy \, dx \pmod{dy}$$

Similarly, as (tangent) vectors are dual to one-forms, we can see that the exterior derivative is the thing you combine with a vector to get a directional derivative.

This is further supported by path integrals; if $\gamma$ is a path from $P$ to $Q$, then $\int_\gamma \, df = f(Q) - f(P)$; so again we see that $df$ is an encoding of how $f$ varies, and the path integral is how we accumulate the variation into a finite difference.

We can argue that $d(df)$ should be zero, ans the variation in the variation of $f$ is second derivative information, and differential forms are only intended to capture first derivative information. Similarly for $df \, df$.

Stokes' theorem expresses the analog of the fundamental theorem of calculus in higher dimensions, giving a way to see the exterior derivative of a differential form as encoding the higher degree variation.

Alternatively, we can appeal to Fubini's theorem to reduce to the one-dimensional case: here's a sketch.

Let x = $(x_1, \ldots, x_n)$ and $dx = dx_1 dx_2 \ldots dx_n$.

Suppose you wish to integrate $$\int_X df \, dx$$ where $X$ is an $(n+1)$-dimensional region. If we let $X_x$ be the one-dimensional region defined by a constant value of $x$, then generalizing Fubini's theorem, we can write this as an iterated integral $$\int_Y \left( \int_{X_x} df \right) dx$$ where $Y$ is some suitable $n$-dimensional space.

The integral $\int_{X_x} df$ is just the alternating sum of the values $f(P)$ where $P$ iterates over the endpoints of the curves comprising $X_x$, where the upper endpoints are added and the lower endpoints are subtracted. It's convenient to write this as an integral over a zero-dimensional surface: $\int_{\partial X_x} f$.

Consequently, the original integral can be written as $$\int_Y \left( \int_{\partial X_x} f \right) dx$$ and again essentially by Fubini's theorem, we can identify this with $$\int_{\partial X} f \, dx$$

Consequently, defining $d(f \, dx)$ as $df dx$ is exactly the right thing to do to generalize the fundamental theorem of calculus to get Stoke's theorem.

Another conceptually nice definition of the exterior derivative is given in Bourbaki (Varietes differentielles et analytiques, Fascicule de resultats), (8.3.4) and (8.3.5). The idea is the following: if $\omega$ is an exterior $p$-form on $X$, consider it as a section $\omega: X \to \Omega^p(X)$ of the bundle $\Omega^p(X)$ of $p$-forms. It makes sense to take its derivative $d\omega$ at each point $x \in X$. Then one sees that $d\omega$ corresponds to a $p+1$ exterior form.

By the way, a natural and simple definition of tangent vector on a smooth manifold is given in the same book in (5.5.1).

• So the derivative of $\omega \colon X \to \Omega^p(X)$ at $x \in X$ is I guess the tangent map $T_x(\omega) \colon T_x X \to T_{\omega(x)} \Omega^p(X)$. How do you get the $p+1$-form? Feb 21, 2013 at 9:55
• That doesn't quite work. Consider applying it to the differential form $x \, dx$ on the real number line, where the differential as a map is not everywhere the same as the differential of the map associated to the zero 1-form. But the exterior derivative is the same. Jun 9, 2020 at 4:52
• This seems to be saying that $\omega$ is a smooth section of the exterior $p$-form bundle $\Omega^p(X)$, $X$ some manifold. Indeed the tangent map is given by $f_*:T_x(\omega): T_x X\to T_{\omega(x)} \Omega^p(X)$ but $d$ doesn't apply to it, $d$ applies the linear map $f^*: A( \Omega^p(X)) \to A(X)$, between the spaces of exterior differential forms, i.e spaces of smooth sections. I think this is how we should get the $p+1$-form. Nov 18, 2020 at 19:07

The exterior derivative is an intrinsic way of talking about the gradient of a function. If you want to understand the intuitive meaning of the exterior derivative of $f$ you should make sure you understand $\nabla f$ properly. I am a little hesitant to post such an answer 5 years into the discussion but as I did not find any occurrence of the string "gradient" on the page I thought this might be useful. In the presence of a metric the relation between them is $\langle \nabla f, V\rangle=df(V)$ for tangent vectors $V$.

• To my mind the gradient only makes sense if you have a metric and may only be compared to $d$ on the level of $0$-forms. So it should not be necessary to understand $\nabla$ properly, before understanding $d$. Jan 18, 2016 at 13:59
• @MichaelBächtold, if one doesn't understand it at the level of $0$-forms one will certainly have trouble understanding it in general. Of course the gradient depends on the metric; that's why it needs to be replaced by the exterior derivative if one wants to work intrinsically. Jan 18, 2016 at 14:04
• I mostly agree with your comment. But I'm not convinced that one should understand $\nabla f$ first. I have the impression that it may even hinder understanding, since $\nabla f$ is usually taught without explaining how it depends on more than $f$. Hence people have to forget hidden assumptions when they try to understand $df$. Jan 18, 2016 at 14:47
• There are some basic elementary geometric facts that form the foundation of understanding here, one of which being that a one-variable function is constant if its derivative is zero, that gradient has to do with direction of steepest descent, etc. Doing research in mathematics is not a formal game but rather involves understanding, without which it is difficult to do research. Jan 18, 2016 at 14:50
• I'm not sure how to interpret your last comment. In case you are suggesting that without a metric the exterior differential necessarily carries no geometric meaning and reduces to its formal rules (Leibniz, $d^2=0$, naturality etc.) then I would disagree. Dan Piponis answer is an attempt to give a geometric understanding without a metric. Another quite elementary and intuitive approach can be found in Anders Kocks book Synthetic Geometry of Manifolds. His approach makes the statement "dual to the boundary operator of singular homology" directly accessible at the infinitesimal level. Jan 18, 2016 at 15:18

First define the exterior derivative for forms defined on an open set $$U \subseteq \mathbb{R^n}$$. This uses the notion of integration of a $$p$$-form over a singular $$p$$-chain, which needs only the integration of $$\mathcal{C}^{\infty}$$-functions over compact subsets of $$\mathbb{R}^p$$ and runs as follows. A singular $$p$$-cube in $$U$$ is a $$\mathcal{C}^{\infty}$$-map $$\sigma : I^p \rightarrow U$$, where $$I := [0,1]$$ is the closed unit interval. Let $$\Omega^p(U) := H^0(U;\wedge^pT^*U)$$ the space of alternating $$p$$-forms on $$U$$; then each $$\omega \in\Omega^p(U)$$ pulls back to a top form $$\sigma^*\omega$$ $$=$$ $$f dx \in \Omega^p(I^p)$$ with $$f \in \mathcal{C}^{\infty}(I^p)$$ and $$dx = dx_1 \wedge \cdots \wedge dx_p$$ the canonical volume element of $$\mathbb{R}^p$$. It thus has an integral $$\int_{\sigma} \omega := \int_{I^p} f dx,$$ and, in fact, this it is what differential forms are made for: born to be integrated.

Next define the vector space of $$p$$-chains to be the free $$\mathbb{R}$$-vector space on the singular $$p$$-cubes, so that a $$p$$-chain $$c_p$$ is a formal linear combination of singular $$p$$-cubes: $$c_p = \sum_{i=1}^k \gamma^i \sigma_i \quad,\quad k\in\mathbb{N}, \gamma \in \mathbb{R}.\tag{1}$$ The integral then extends to $$p$$-chains by linearity; $$\int_{c_p}\omega := \sum_{i=1}^k \gamma^i \int_{\sigma_i} \omega.$$

As a next ingredient we need that any $$p$$-chain $$c_p$$ has a boundary $$\partial c_p$$ which is a $$(p-1)$$-chain. We first define it on singular $$p$$-cubes $$\sigma$$ by $$\partial \sigma := \sum_{j=1}^p (-1)^j (\sigma \circ d^j_- - \sigma \circ d^j_+),$$ where the singular $$(p-1)$$-cubes $$d^j_{\mp}$$ in $$I^p$$ define the $$j$$-th front and back boundary faces: $$d^j_-(x^1, \dots, x^{p-1}) := (x^1, \dots, x^j, 0, x^{j+1}, \dots, x^{p-1}),$$ $$d^j_+(x^1, \dots, x^{p-1}) := (x^1, \dots, x^j, 1, x^{j+1}, \dots, x^{p-1}).$$ We extend this boundary operator to $$p$$-chains by linearity: $$\partial c_p := \sum_{i=1}^k \gamma^i \partial \sigma_i$$ with $$c_p$$ given by (1).

As a last ingredient we need that a point $$P \in U$$ and a $$p$$-tuple of vectors $$X:= (X_1, \dots, X_p)$$ (viewed as tangent vectors at $$P$$ to $$U$$) define a singular $$p$$-chain $$[X]_P : I^p \rightarrow U$$ via $$[X]_P(x^1, \dots, x^p) := P+\sum_{k=1}^p x^k X_k$$ as soon as the $$X_k$$ are so small that all the $$P+x^kX_k$$ are in $$U$$ for all $$k$$. In fact, these simple linear singular chains are all what is needed of this formalism to define the exterior derivative, to which we proceed next.

After this preliminaries, we now want, given $$\omega \in \Omega^p(U)$$, define its exterior derivative $$d\omega \in \Omega^{p+1}(U)$$. We do this pointwise at any point $$P \in U$$ by exhibiting the value $$d\omega_P$$, as an alternating $$(p+1)$$-form, takes on any $$(p+1)$$-tuple of (tangent) vectors $$(X_1, \dots, X_{p+1})$$ $$\in$$ $$(\mathbb{R^n})^{p+1}$$. We define $$\fbox{d\omega_P(X_1, \dots, X_{p+1}) := \lim_{t \rightarrow 0} \dfrac{1}{t^{p+1}} \int_{\partial([tX]_P)} \omega.}$$

Finally, for the general case of the exterior derivative of a $$p$$-form $$\omega$$ on an $$n$$-dimensional manifold $$M$$, just take charts $$\phi: V \rightarrow U$$ with $$V$$ open in $$M$$, $$U$$ open in $$\mathbb{R}^n$$, with the $$V$$ covering $$M$$, and put $$(d\omega)|V := \phi^*d\eta \quad \text{with}\quad \eta := (\phi^{-1})^*(\omega|V) \in \Omega^p(U).$$ The transformation formula for multivariate integrals then shows that the $$(d\omega)|V$$ glue well on the overlaps, thus yielding a global well-defined $$d\omega$$.

Loosely speaking, this defines the exterior derivative as a "volume derivative", a flux density through the boundary of an infinitesimal $$(p+1)$$-dimensional parallelepiped and so has as a built-in an infinitesimal version of Stokes' Theorem.

• This is similar in parts to the answer by Petya, although you have included more details. Aug 5, 2020 at 2:09