What does the adjective "natural" actually mean? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:40:11Z http://mathoverflow.net/feeds/question/56938 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean What does the adjective "natural" actually mean? unknown (google) 2011-02-28T23:49:35Z 2012-09-13T20:18:25Z <p>Terms like "in the natural way" or "the natural X" are used frequently in mathematical writing. While it is certainly clear most of the time what is meant, on occasion, I have been confounded. The word "natural" seems to be one of the most ambiguous terms used in formal mathematics. I have never seen anyone actually define it. People just use it and expect others to understand it.</p> <p>What exactly is meant by "natural" in mathematical writing?</p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/56943#56943 Answer by Daniel Miller for What does the adjective "natural" actually mean? Daniel Miller 2011-03-01T00:01:49Z 2011-03-02T19:53:51Z <p>I'm not sure if this is what you are looking for, but the word "natural" is often used to imply that the object being discussed is in fact a <a href="http://en.wikipedia.org/wiki/Natural_transformation" rel="nofollow">Natural Transformation</a>. For example, the isomorphism $V\cong V^{**}$ between a finite dimensional vector space and its double dual is natural in the sense that it is a natural isomorphism between functors. The same holds for the isomorphism $G/\mbox{Ker}\:\phi\cong H$ when $\phi:G\to H$ is a surjective homomorphism and $G,H$ are groups.</p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/56946#56946 Answer by quid for What does the adjective "natural" actually mean? quid 2011-03-01T00:18:54Z 2011-03-01T00:18:54Z <p>Typically, there is no exact meaning, beyond 'reasonable or expected in a particular situation' as in plain English.</p> <p>One thing that is close to giving a precise meaning to 'natural' is given in Daniel Miller's answer.</p> <p>However, not at all every usage of 'natural' in mathematical writing falls into this category. So, the situation is just as you describe: the choice or the way was the only one that seemed reasonable to the author and thus it was the natural one. And, yes, sometimes when reading it this can be frustrating. Though, I should add, that avoiding to write it (and similar things) can be really hard.</p> <p>Assuming you are not yet reading mathematics for too long a time, let me end this, in some way unhelpful, answer by mentioning that over time one develops the same or a similar sense of 'reasonable and expected' as most authors (at least in ones field of expertise) and thus the situation (at least subjectively) becomes considerably better over time. </p> <p>Note: I am a different unknown. </p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/56948#56948 Answer by John Sidles for What does the adjective "natural" actually mean? John Sidles 2011-03-01T00:40:01Z 2012-09-13T20:18:25Z <p><strong>Edit</strong>&nbsp; As an example of the increasing prevalence of the notion of <em>naturality</em> in contemporary mathematics, it is notable that Shinichi Mochizuki’s <a href="http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture" rel="nofollow">four preprints</a> asserting a proof of the ABC conjecture employ the word "natural" and its derivatives on more than six hundred occasions (for details and several related quotations, see <a href="http://rjlipton.wordpress.com/2012/09/12/the-abc-conjecture-and-cryptography/#comment-26756" rel="nofollow">this post</a> on <em>Gödel's Lost Letter and P=NP</em>).</p> <hr> <p>In accord with Daniel Miller's answer above, the study of "naturality" as a formal abstraction in mathematics can be traced back largely to two articles by Saunders Mac Lane and Samuel Eilenberg: <i>Natural isomorphisms in group theory</i> (1942) and <i>General theory of natural equivalences</i> (1945). Both articles are well worth reading.</p> <p>How did these ideas arise? Roughly speaking, Eilenberg and MacLane began by recognizing that if an arbitrary choice of coordinates makes a difference to the quantities you are calculating and/or the theorems you are proving, then those quantities and theorems are not natural. Mac Lane has described this process in <i>The development and prospects for category theory</i> (1996) as follows:<blockquote>I emphasize that the notions category and functor were not formulated or put in print until the idea of a natural transformation was also at hand.</blockquote>Thus, one good start for students is to acquire a thorough practical grasp of naturality in the context of coordinate transformations in linear algebra and differential geometry. </p> <p>A concrete example of a canonical usage of "naturality", which is accompanied by an extensive motivating discussion, is given in John Lee's <em><a href="http://books.google.com/books?id=eqfgZtjQceYC&amp;pg=PA310" rel="nofollow">Introduction to Smooth Manifolds</a></em> as the following lemma:</p> <blockquote> <p>Lemma 12.16: (Naturality of the Exterior Derivative) </p> <p>If $G\colon M\to N$ is a smooth map, then the pullback map $G^\star\colon \mathcal{A}^k(N)\to \mathcal{A}^k(M)$ commutes with $d$. That is, for all $\omega \in \mathcal{A}^k(N)$, we have $G^\star(d\omega) = d(G^\star\omega)$.</p> </blockquote> <p>As a commutative diagram the above lemma exhibits a canonical form:</p> <p><code>$$\begin{array}{c@{}ccc@{}c} &amp;&amp;d&amp;&amp;\\ &amp;\mathcal{A}^k(N)&amp;\longrightarrow&amp;\mathcal{A}^{k+1}(N)\\[2ex] G^\star\!\!\!\!\!\!&amp;\big\downarrow&amp;&amp;\big\downarrow&amp;\!\!\!\!\!\!\!G^\star\\[2ex] &amp;\mathcal{A}^k(M)&amp;\longrightarrow&amp;\mathcal{A}^{k+1}(M)\\ &amp;&amp;d&amp;&amp; \end{array}$$</code></p> <p>Being interested in practical applications of geometrically "natural" formalisms for <a href="http://mathoverflow.net/questions/54232/a-book-you-would-like-to-write/54364#54364" rel="nofollow">quantum systems engineering</a>, I've studied the usage in MacSciNet reviews of the words "natural*" (chiefly "natural" and "naturality") and "universal*" (chiefly "universal" and "universality"). </p> <p>Here are the numbers; their use is burgeoning! </p> <ul> <li>Year-Range (natural*, universal*) </li> <li>2001-2005: (16788 uses, 05288 uses)</li> <li>1996-2000: (14880, 04977)</li> <li>1991-1995: (12550, 04432)</li> <li>1986-1990: (10335, 03343)</li> <li>1981-1985: (08775, 03013)</li> <li>1976-1980: (07402, 02412)</li> <li>1971-1975: (05668, 02040)</li> <li>1966-1970: (03466, 01167)</li> <li>1961-1965: (02211, 00610)</li> <li>1956-1960: (01368, 00406)</li> <li>1951-1955: (00880, 00253)</li> <li>1946-1950: (00502, 00107)</li> <li>1941-1945: (00251, 00060)</li> </ul> <p>So to judge by the literature, it seems that we are entering into a Golden Era of mathematical "naturality" and "universality" ... we can hope so, anyway! :)</p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/56954#56954 Answer by Jeff Strom for What does the adjective "natural" actually mean? Jeff Strom 2011-03-01T01:02:56Z 2011-03-01T01:02:56Z <p>According to Mac Lane (as I remember it from Categories for the Working Mathematician), Eilenberg and Mac Lane invented categories so they could talk about functors, and they wanted to talk about functors so they could define "natural."</p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/56956#56956 Answer by Sándor Kovács for What does the adjective "natural" actually mean? Sándor Kovács 2011-03-01T01:38:28Z 2012-02-23T19:09:15Z <p>Actually, there is an exact meaning, but it is not always used in that sense. For two functors $\mathsf F,\mathsf G:\mathscr A\to \mathscr B$ a <em>natural transformation</em> is a morphism of functors $\eta:\mathsf F\to\mathsf G$ that is compatible with the functors in the obvious (sic!) way.</p> <p>For instance if $\mathsf F={\rm id}$ is the identity and $\mathsf G=(\underline{\quad} )^{*}$ is the dual of <em>finite-dimensional</em> vector spaces, then even though $\mathsf F(V)\simeq \mathsf G(V)$, there is no natural transformation between $\mathsf F$ and $\mathsf G$ that gives this isomorphism. On the other hand $\mathsf F$ is naturally isomorphic to $\mathsf D:=\mathsf G\circ\mathsf G$ via the natural transformation induced by the usual map to the double dual.</p> <p>Of course, often people say "there is a natural choice of" whatever. That usually means that the "choice" actually does not involve a "choice". In other words, two different people would be expected to make the same choice.</p> <p>There is however a danger involved that some authors overlook. There are situations when there are more than one natural choices to make. In other words, just because two choices are both natural, one should not assume that they are necessarily the same. For instance, often one can end up with $(-1)$-times the other choice (say for a map) and then whether they are equal or their sum is zero makes a big difference. For an example, consider choosing a generator of the infinite cyclic group, a.k.a., $(\mathbb Z,+)$. One might be led to believe that "the" natural choice of a generator for $(\mathbb Z,+)$ is $1\in \mathbb Z$, but as long as this is only a <em>group</em> there is no way to distinguish $1\in \mathbb Z$ and $-1\in \mathbb Z$. They both generate the group and they are each other's inverses. In other words, there are two <em>natural</em> choices of a generator of the <em>group</em> $(\mathbb Z,+)$. Of course, once it is a ring, then $1\in \mathbb Z$ is the unity, while $-1\in \mathbb Z$ is not, and there is only one choice for the unity, but actually then the question of this choice being natural is moot since there is only one choice at all, so it's kind of silly to say it is natural.</p> <p><strong>Addendum</strong> (to answer the question raised in the comment below by <em>unknown</em>)</p> <p>A two dimensional vector space does not have a "natural" inner product. A two dimensional vector space <em>with a chosen basis</em> does: If $V$ is a vector space (over the field $k$) with basis <code>$\mathbf v_1,\mathbf v_2\in V$</code> then one can define a "natural" inner product by <code>$&lt;\mathbf a,\mathbf b&gt;:=a_1b_1+a_2b_2$</code> where $\mathbf a=a_1\mathbf v_1+a_2\mathbf v_2$ and $\mathbf b=b_1\mathbf v_1+b_2\mathbf v_2$. But this is only natural <em>after</em> the basis has been chosen. Basically the problem is that the definition of the inner product depends on the <em>choice</em> of the basis, so this definition is only natural if there is a natural choice of a basis (assuming there is no other structure present that could give a natural inner product; for more on this see the example of the wedge product below). </p> <p>For instance if $V$ is given with a basis as above, then there is a natural choice of a basis (called the <em>dual basis</em>) for $V^*$ the dual space of $V$: Let $\phi_1: V\to k$ be defined by $\phi_1(\mathbf v_1)=1$, $\phi_1(\mathbf v_2)=0$ and extended by linearity and similarly $\phi_2: V\to k$ be defined by $\phi_2(\mathbf v_1)=0$, $\phi_2(\mathbf v_2)=1$ and extended by linearity. So, <em>if</em> you already have a chosen basis on $V$ you can find a natural choice of a basis on $V^*$ and with that you can find a natural choice of an inner product, but this will not be a natural choice if you consider the vector space without the given basis.</p> <p>Otherwise, I agree, it is hard to tell what people mean by "natural". As said by many people in various answers, the essence is whether you can do the construction without making a sort of a random choice when choosing a different element would be equally good. In this example, to give an inner product you need to give a basis and unless you have some extra structure, choosing any given basis is equal to choosing any other, so the choice is non-natural. </p> <p>On the other hand if there is some extra structure on the vector space then there may be a natural choice for an inner product, or more generally for a bilinear form. For instance if your vector space is a space of differential forms on a manifold, then there is no natural choice of basis, but there is a natural choice of an alternating bilinear form: the wedge product. This is actually pretty good, because this means that it is possible to define this on a manifold <em>locally</em>: picking a chart is a non-natural choice and defining the wedge product of two differential forms locally seems like it depends on a lot of choices, but it ends up being independent of these in the sense that choosing a different chart you get the same wedge product it just looks different because it is in a different basis.</p> <p><strong>Addendum 2</strong> (I realized that this might be an interesting comment while writing this <a href="http://mathoverflow.net/questions/57508/unique-factorisation-and-the-fact-that-mathbb-a2-0-is-not-an-affine-variety/57594#57594" rel="nofollow">answer</a> to another MO question).</p> <p>Here is an example of the importance of naturality: Suppose $M$ is a manifold and $U\subseteq M$ and open set. Then there is a <em>natural</em> homomorphism from the ring of regular functions on $M$ to the ring of regular functions on $U$. (If you like adjust "regular" for your favourite category; continuous, smooth, holomorphic, etc.). It can happen that this homomorphism is an isomorphism and then it has nice consequences. However, it is often important that in order to get the nice consequence one needs that <em>the</em> homomorphism induced by the embedding is an isomorphism and not that there exists <em>some</em> isomorphism. In other words, one could say that <em>the</em> natural homomorphism (=the one induced by the embedding $U\subseteq M$) is an isomorphism. For an explicit example where this matters see <a href="http://mathoverflow.net/questions/57508/unique-factorisation-and-the-fact-that-mathbb-a2-0-is-not-an-affine-variety/57594#57594" rel="nofollow">the above mentioned answer</a>.</p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/56967#56967 Answer by Neil Strickland for What does the adjective "natural" actually mean? Neil Strickland 2011-03-01T05:28:16Z 2011-03-01T05:28:16Z <p>As several people have mentioned, there is a precise technical meaning in category theory. This does not fully capture the informal meaning. For example, consider the category $\mathbf{Groups}$ of groups and homomorphisms, and write $Z(G)$ for the centre of $G$. This does not give a functor $\mathbf{Groups}\to\mathbf{Groups}$ (because a homomorphism $\phi:G\to H$ need not carry $Z(G)$ into $Z(H)$), but it is certainly a natural construction in an informal sense. However, we can consider the subcategory $\mathbf{Groups}'$ of groups and isomorphisms, and then $Z$ will give a functor $\mathbf{Groups}'\to\mathbf{Groups}$. This kind of categorical naturality with respect to isomorphisms is much closer to the informal meaning. I would say that any informally natural construction is categorically functorial/natural with respect to isomorphisms, but there are counterexamples to the converse. For example, there is a forgetful functor $U:\mathbf{Rings}\to\mathbf{Sets}$ and a transformation $\alpha:U\to U$ given by </p> <p>$$\alpha_R(a) = 182567 a^{46} - 17576 a^9$$</p> <p>This is a natural transformation in the sense of category theory, but obviously involves choices that are essentially arbitrary. </p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/56999#56999 Answer by Anixx for What does the adjective "natural" actually mean? Anixx 2011-03-01T13:42:21Z 2011-03-01T13:42:21Z <p>I perceive that a "natural" solution to a problem differs from other solutions in that in addition to all the properties of other solutions, it also satisfies some additional requirements which either particularly useful for the solution's purpose or greatly simplify the expressions.</p> <p>Just for example, when one speaks about interpolation of a function over a number of points, the "natural" solution will differ from the infinitely many other solutions in that it will also satisfy certain functional equations of the original function not only on the domain of definition of the original function, but also in the intermediate points. Thus one can speak about natural generalization/interpolation of the function. </p> <p>For example, Gamma function is a natural generalization of factorial (up to a shift) because in addition to satisfying the functional equations of factorial in integer points it also satisfies the same equations in other points and has other useful properties. </p> <p>The exponential function with base e is also natural in that in addition to having all properties of exponential function it also has a unique property of being its own derivative.</p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/57083#57083 Answer by Amritanshu Prasad for What does the adjective "natural" actually mean? Amritanshu Prasad 2011-03-02T07:33:59Z 2011-06-13T04:55:21Z <p>The sense in which "natural" is used by many modern mathematicians (as coming from natural transformation of functors) is different from "canonical" (which means that it does not involve making any choices beyond the data given). The fact that both these adjectives often apply to the same object (e.g., the embedding of a vector space in its double dual), should not serve as an excuse to confuse their meanings-something can be natural without being canonical, and something can be canonical without being natural, as I shall illustrate with examples from group theory.</p> <p>For the record, here is the official definition of a natural transformation. </p> <blockquote> <p>If $\mathcal C$ and $\mathcal D$ are categories, $F_1$ and $F_2$ are functors $\mathcal C\to \mathcal D$, then a natural transformation $F_1\to F_2$ is a family of morphisms <code>$N(X):F_1(X)\to F_2(X)$</code> in $\mathcal D$ (one for each object $X$ of $\mathcal C$) such that for every morphism $f:X\to Y$ in $\mathcal C$, <code>$N(Y)\circ F_1(f)=F_2(f)\circ N(X)$</code>.</p> </blockquote> <p>It should be noted that natural applies to a <em>family</em> of morphisms, not to a single morphism.</p> <h2>The groupoid associated to a group</h2> <p>Let $G$ be a group. Let $\mathcal C$ be the category with one object $*$, with <code>$\mathrm{Hom}(*,*)=G$</code>, composition being determined by the group law. This category $\mathcal C$ is known as the <em>groupoid</em> associated to $G$. Note that</p> <ul> <li>A functor <code>$F:\mathcal C\to \mathcal C$</code> is determined by what it does to morphisms (at the level of objects, it just takes <code>$*$</code> to <code>$*$</code>). One easily checks that the monoid of functors <code>$\mathcal C\to \mathcal C$</code> is isomorphic to the monoid of endomorphsims $G\to G$.</li> <li>If $F_1$ and $F_2$ are two functors <code>$\mathcal C\to \mathcal C$</code> (identify them with endomorphisms of $G$) then a morphism $g:*\to *$ (identify it with an element of $G$) is a natural transformation if and only if <code>$g\circ F_1(h)=F_2(h)\circ g$</code> for all $h\in G$.</li> </ul> <h2>Natural, but not canonical</h2> <p>The natural endomorphisms of the identity functor $\mathcal C\to \mathcal C$ are the morphisms in the center of $G$. Only the identity element here could be considered canonical. The others would be non-canonical, at least until more data is provided.</p> <h2>Canonical, but not natural</h2> <p>If $F_1$ and $F_2$ are two distinct functors $\mathcal C\to \mathcal C$ (group homomorphisms $G\to G$), then the identity morphism $*\to *$ (which is indisputably canonical) is not a natural transformation $F_1\to \mathcal F_2$.</p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/102705#102705 Answer by robot for What does the adjective "natural" actually mean? robot 2012-07-19T23:56:20Z 2012-07-19T23:56:20Z <p>There is a book called <a href="http://www.emis.de/monographs/KSM/index.html" rel="nofollow">Natural operations in differential geometry</a>, which as far as I know captures what natural means for a differential geometer. Natural bundles are defined basically as functors from the category of manifolds with local diffeomorphisms to the category of fibered manifolds. Natural operator between natural bundles is then a local operator which commutes with local diffeomorphisms.</p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/102726#102726 Answer by Per Alexandersson for What does the adjective "natural" actually mean? Per Alexandersson 2012-07-20T08:46:53Z 2012-07-20T08:46:53Z <p>For me, a natural constraint on some parameters in a theorem, is that without these constraints, either the theorem is false for obvious reasons, or trivial on the complement of the constraints but the proof method needs these constraints.</p> <p>For example, in many theorems, it is natural to consider only odd primes.</p> http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/102741#102741 Answer by Lee Mosher for What does the adjective "natural" actually mean? Lee Mosher 2012-07-20T13:32:56Z 2012-07-20T13:32:56Z <p>One issue I have with all of the answers is that mathematical writing is writing by humans. Much of the language in a mathematical paper written by a human is plain language, intended to impart understanding from one human to another human, rather than formal language intended to codify a proof. In proving many existence theorems, one has to make a rather wild guess at the definition of the object whose existence is asserted, long before one is able to carry out the actual proof that the object satisfies all of the required properties of the theorem. Such wild guesses have many different motivations in humans. I often use the word "natural" to let my reader know what motivated me to guess at the right object. Sometimes the motivation was background knowledge or experience, sometimes it was just blind intuition which I cannot account for, and sometimes it is motivated by what felt "natural". I would not assert that there is any formal mathematical definition for this use of "natural". I suppose we'll have to wait to see what happens with this term once mathematical papers are written by computers to impart understanding from one computer to another.</p>