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I would like to ask if anyone could share any specific experiences of discovering nonequivalent definitions in their field of mathematical research.

By that I mean discovering that in different places in the literature, the same name is used for two different mathematical objects. This can happen when the mathematical literature grows quickly and becomes chaotic and of course this could be a source of serious errors. I have heard of such complaints by colleagues, mostly with respect to definitions of various spaces and operators, but I do not recall the specific (and very specialised) examples.

My reason for asking is that I'm currently experimenting with verification using proof assistants and I'd like to test some cases that might be a source of future errors.

EDIT: Obviously, I would be interested to see as many instances of this issue as possible so I would like to ask for more answers. Also, it would be even more helpful if you could provide references.

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    $\begingroup$ A classic example is "compact", which can mean either "Hausdorff and every open cover has a finite subcover", or just "every open cover has a finite subcover". $\endgroup$
    – Arturo Magidin
    Commented Nov 22, 2017 at 19:12
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    $\begingroup$ @ArturoMagidin This leads to the classic dialogue "The space $X$ is quasicompact, so..." "What does quasicompact mean?" "It means compact." $\endgroup$
    – Will Sawin
    Commented Nov 22, 2017 at 19:35
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    $\begingroup$ In my experience at least, the only people who think "compact" includes Hausdorff are algebraic geometers. I've never heard a topologist use "compact" to mean anything other than "every open cover has a finite subcover". $\endgroup$
    – John Pardon
    Commented Nov 23, 2017 at 3:29
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    $\begingroup$ @JohnPardon: there is also a cultural thing here. In France, it is standard to include "Hausdorff" in the definition of a compact space. $\endgroup$
    – Taladris
    Commented Nov 23, 2017 at 14:49
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    $\begingroup$ @Taladris, maybe in france, everyone's an algebraic geometer $\endgroup$
    – Vivek Shende
    Commented Nov 29, 2017 at 0:54

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Perhaps the mother of all examples is "natural number". You can start an internet flame war by asking whether zero is a natural number.

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    $\begingroup$ @July, so is $\mathbb{N}^+$ not a waste of notation? :P $\endgroup$
    – Peter Taylor
    Commented Nov 23, 2017 at 11:20
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    $\begingroup$ @July this will be my last comment on the subject. The notation $\mathbb{N}_0$ does not make me happy at all; it is far too cumbersome for such a basic object. If the set including zero is a more important mathematical object, then it deserves the simpler notation. The word "natural", when used in mathematics, should refer to naturalness in mathematics; there's no reason a historical accident of terminology should guide the choice of appropriate notation; and I dispute that there's anything less natural about "the number of elephants in this room" than "the number of fingers on my hand". $\endgroup$
    – Mike Shulman
    Commented Nov 23, 2017 at 17:44
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    $\begingroup$ I like to write $\Bbb{Z}_{>0}$ and $\Bbb{Z}_{\ge0}$. Upside: impossible to misinterpret, doesn't annoy people by picking favourites between the two. Downside: a little clunky. $\endgroup$
    – Gareth McCaughan
    Commented Nov 24, 2017 at 23:16
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    $\begingroup$ It turns out that one can start an internet flamewar merely by mentioning that one can start an internet flamewar by asking whether zero is a natural number. $\endgroup$
    – David Richerby
    Commented Nov 25, 2017 at 12:28
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    $\begingroup$ I am OK with folks who support the idea that zero is not a natural number. All I ask is they be consistent in their notation and write $\mathbb{N}=\{I, II, III, IV, V, VI, VII, VIII, IX, X,...\}$ $\endgroup$
    – Abdelmalek Abdesselam
    Commented Nov 26, 2017 at 21:14
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Linear functions:

In high-school algebra (sometimes called "pre-calculus"), we are taught that linear functions are those of the form $y=mx+b$, because they are graphed by a straight line in the plane.

Then we study linear algebra in university, and realize that for a function $f$ to be linear it must satisfy $f(0)=0$, as a special case $a=0$ of the linearity requirement that $f(ax)=af(x)$, and the functions that were called "linear" in high school are not really linear (unless $b=0$) but affine.

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    $\begingroup$ I teach my algebra students the term affine, but let them know that in most cases they are called "linear" despite not actually being linear. $\endgroup$
    – Alex Jones
    Commented Nov 23, 2017 at 8:40
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    $\begingroup$ Computer science also uses "linear" to mean "cost grows directly proportional to problem size, ignoring a constant fixed cost", which is affine. $\endgroup$
    – Eric Lippert
    Commented Nov 24, 2017 at 15:37
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    $\begingroup$ @EricLippert: They're only interested in the asymptotic behaviour when $x$ gets to infinity. That's why $b$ is ignored. The number and speed of CPU is arbitrary, so $a$ is set to $1$. $\endgroup$
    – Eric Duminil
    Commented Nov 24, 2017 at 16:35
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    $\begingroup$ @EricLippert Computer scientists would also call $n + \log{n}$ 'linear', even though it's not even affine. $\endgroup$
    – Adam P. Goucher
    Commented Dec 20, 2017 at 17:08
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    $\begingroup$ An ODE $\dot u=Au+b$ is still called a "linear equation", and $\dot u=Au$ "linear homogeneous" $\endgroup$
    – Pietro Majer
    Commented Jan 22, 2018 at 22:16
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Not a word but a piece of notation: Sometimes I have seen $\subset$ used to mean "is a proper subset of" while other times I have seen it used to mean "is a subset of".

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    $\begingroup$ In my university this is quite an interesting example: my first course there was an introductory notation and structures course, introducing $\subset$ as proper subset. Then literally every other teacher proceeded to use $\subset$ as a generic subset. $\endgroup$
    – tomsmeding
    Commented Nov 23, 2017 at 8:12
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    $\begingroup$ That is why I always use $\subseteq$ and $\subsetneqq$... $\endgroup$
    – Dirk
    Commented Nov 23, 2017 at 10:31
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    $\begingroup$ @DirkLiebhold As an alternative to the latter there are also the (at least to me) aesthetically more pleasing $\varsubsetneq$ and $\subsetneq$ $\endgroup$
    – Jules Lamers
    Commented Nov 23, 2017 at 12:48
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    $\begingroup$ @JulesLamers one careless stroke in handwriting or rubbed off fiber of very worn out paper, and your notation confuses the reader. $\endgroup$
    – Ruslan
    Commented Nov 24, 2017 at 10:13
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    $\begingroup$ Interestingly, I have never seen the symbol $<$ used to mean "less than or equal to" nor have I seen anyone feel compelled to use the analogue of Dirk Liebhold's solution (by inventing a symbol for "less than but not equal to"). I wonder why the two notations have evolved differently even though they seem extremely closely related. Maybe "proper subset" just doesn't come up that often. $\endgroup$
    – Timothy Chow
    Commented Nov 24, 2017 at 15:05
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Everyone knows what a curve is, until he has studied enough mathematics to become confused through the countless number of possible exceptions. — F. Klein

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    $\begingroup$ Of course, sometimes a curve is a map from $\mathbb{R}$ into a manifold, and other times it is a Riemann surface, and still other times it is a reduced 1-dimensional separated scheme of finite type over a field. $\endgroup$
    – Robert Furber
    Commented Nov 23, 2017 at 14:37
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    $\begingroup$ Heck, people keep telling me $\mathbb{C}$ is a line. $\endgroup$
    – Eric Towers
    Commented Nov 24, 2017 at 6:32
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    $\begingroup$ I remember being confused about how some algebraic geometers and number theorists were using the word "curve" until I realized that I was tacitly (and incorrectly) assuming that a "curve" was what, in their language, was a "set of $F$-rational points" of the curve. It takes a while to unlearn the idea that a curve is a set of points. $\endgroup$
    – Timothy Chow
    Commented Nov 24, 2017 at 15:21
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    $\begingroup$ @TimothyChow, which is funny, because it is undoing the complicated process (foreign to Euclid, for example) of first learning that a curve is a set of points …. $\endgroup$
    – LSpice
    Commented May 25, 2019 at 16:55
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    $\begingroup$ @TimothyChow, good point; but (at the risk of de-railing a years-old comment section with this) I feel that, with the idea of tracing out a curve as the locus (of points, I can't help saying) satisfying a certain condition being so familiar to him, Euclid would have been much more comfortable with the algebraic-geometry than with the point-set approach to curves. $\endgroup$
    – LSpice
    Commented May 26, 2019 at 20:00
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Another kind of answer. There is

increasing; strictly increasing

and there is

nondecreasing; increasing

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    $\begingroup$ Reminiscent of the basic definitions in Jacobson's "Lie Algebras", where he says "A nonassociative algebra is [doesn't mention associativity]"; and $\endgroup$
    – Jeff Strom
    Commented Nov 23, 2017 at 14:45
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    $\begingroup$ ... "an associative algebra is a nonassociative algebra that [is associative]" $\endgroup$
    – Jeff Strom
    Commented Nov 23, 2017 at 14:45
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    $\begingroup$ There is a nice alternative: $\le$-preserving vs $<$-preserving. $\endgroup$
    – Tom Ellis
    Commented Nov 23, 2017 at 15:08
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    $\begingroup$ And "increasing" can also mean: $x < f(x)$. $\endgroup$
    – Gerald Edgar
    Commented Nov 23, 2017 at 17:18
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    $\begingroup$ @JeffStrom In the same way that a manifold is a manifold with boundary without boundary? $\endgroup$
    – Neal
    Commented Nov 25, 2017 at 15:52
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Polygons!

Is a polygon a sequence of vertices together with the edges that connect consecutive vertices? If so, can two distinct vertices be the same point? How about two consecutive vertices? Or a pair of vertices two indices apart: can we go from A, to B, and then directly back to A?

Or if not, can the edges of a polygon intersect each other, aside from the necessary intersections between consecutive edges at vertices?

Can we say that points are inside or outside the polygon? If so, and the polygon is self intersecting, do the points within the intersection cancel, or stay, or get counted with multiplicity?

Or is a polygon an area whose boundary consists of line segments? If so, can its boundary be disconnected (can It have holes)? Can it pinch to a single point like a figure eight?

Are the vertices ordered? Or at least ordered cyclically? Are the edges directed? If so, do the directions of consecutive edges need to match up?

For that matter, do the edges need to connect up at all, or can our polygon have a starting point and an ending point with no edge connecting them?

And that's not even getting into polyhedra...

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    $\begingroup$ Does every polygon in a space necessary lie in a plane? $\endgroup$
    – Andrei Smolensky
    Commented Nov 23, 2017 at 11:57
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    $\begingroup$ I participated in a consortium that has the task of defining rules for a map format for vehicle routing and one of problems was to define, what kind of planar polygons are allowed; trying to work out mathematically sound mandatory rules was a major pain because different participants had different ideas of what is a simple polygon (should self-touching be allowed and if yes, what kind of restrictions should be imposed on edge overlaps) $\endgroup$
    – Manfred Weis
    Commented Nov 26, 2017 at 11:24
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    $\begingroup$ Does a polygon have finitely many vertices? $\endgroup$
    – Christopher King
    Commented Jan 22, 2018 at 22:37
  • $\begingroup$ @AndreiSmolensky what kind of plane? $\endgroup$
    – Christopher King
    Commented Jan 22, 2018 at 22:38
  • $\begingroup$ For polyhedra, we have Lakatos' whole book of course. $\endgroup$
    – Balazs
    Commented Sep 19, 2023 at 9:37
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Positive definite matrix (and related terms). Most authors require these to be hermitian (or symmetric in the real case), but not all.

EDIT: also, "positive" can be ambiguous even for numbers: most authors, especially in English, consider $0$ to be neither positive nor negative, but according to Bourbaki it is both, and many authors (especially French ones) follow Bourbaki's convention, using "strictly positive" or "strictly negative" for what most authors would call "positive" or "negative".

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    $\begingroup$ Many prefer the distinction between "positive" vs. "non-negative", which clears possible confusion. "Strictly positive" is more of emphasizing the positiveness so as to avoid any potential source of confusion. Using both "positive" and "strictly positive" is kinda weird. $\endgroup$
    – M.G.
    Commented Nov 22, 2017 at 22:25
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    $\begingroup$ Functional analysis generally uses "positive" to mean "nonnegative" because the notion of strictly positive does not generalize well to infinite dimensional spaces (the interior of the positive cone of an ordered Banach space can be empty). Functional analysts also call "nonexpansive" mappings "contractions". It makes sense that the more frequently used concept gets the shorter name. $\endgroup$
    – Robert Furber
    Commented Nov 23, 2017 at 13:35
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    $\begingroup$ @July and "not non-negative" is not equivalent to "negative." You get other issues. $\endgroup$
    – Clement C.
    Commented Nov 24, 2017 at 3:13
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    $\begingroup$ @July "by contradiction, suppose f is not non-negative [...]" $\endgroup$
    – Clement C.
    Commented Nov 24, 2017 at 15:05
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    $\begingroup$ @July I think there are far fewer issues with "positive, strictly positive" than with "non-negative, positive". After all we have "contains, strictly contains", and nobody considers that kinda weird. Trying to weaken a notion by negating its opposite is a bad idea in general, because it depends on just how the opposite is taken, and may not give the right notion anyway. See math.stackexchange.com/a/115951/18880 $\endgroup$
    – Marc van Leeuwen
    Commented Dec 1, 2017 at 9:06
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Tensor: for some people, a tensor is an element of a tensor product of vector spaces. For others it’s a section of a tensor product of tangent and cotangent bundles on a manifold. Members of the first group would call that latter notion a tensor field.

More interestingly:

Tensor rank: a tensor of rank $r$ is either a sum of $r$ simple tensors (outer products) or an element of a tensor product of $r$ vector spaces (an “$r$-dimensional array of numbers”). For example, a matrix is a rank $2$ tensor in the latter sense.

I speculate that at least 25% of Math Stack Exchange questions about tensors are about confusion over one or both of the above.

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    $\begingroup$ I was half expecting the physicist's "definition" here: "A tensor is an object that transforms like a tensor." (Student: "?!" Professor: "It's basically anything with indices on it"). ;) $\endgroup$
    – balu
    Commented Nov 23, 2017 at 20:27
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    $\begingroup$ @balu Followed up by "oh, so like the Christoffel symbols?" "No." $\endgroup$
    – A. Thomas Yerger
    Commented Nov 25, 2017 at 8:47
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    $\begingroup$ So the zero matrix would have rank $2$. Interesting :P $\endgroup$
    – Qfwfq
    Commented Apr 28, 2018 at 13:19
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    $\begingroup$ Maybe it would be better to say "tensor of degree $r$" to mean an element of $V^{\otimes r}$, which is consistent with the fact that the tensor algebra is a graded algebra so its elements have a degree. $\endgroup$
    – Qfwfq
    Commented Apr 28, 2018 at 13:21
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    $\begingroup$ Let's not forget the machine learning definition of tensor: a multidimensional array (no further properties required). $\endgroup$
    – littleO
    Commented Apr 28, 2018 at 16:46
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"Function", prior to about 1910 always meant the $y$ in $y=f(x)$ (Look up any definition form that period). Since roughly 1920 it's officially the $f$. Physicists, engineers and many applied mathematicians still mean $y$ when they talk about functions today. But, since the hijacking of the term, they lack adequate terminology to make the distinction between $y$ and $f$. (Probably they are not even aware of the distinction, due to the sloppy use of the word function in most calculus textbooks combined with the common notation $y=y(x)$.)

More on the history of this strange development can be found here and here.

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    $\begingroup$ People who regard $y$ in $y = f(x)$ as a function actually implicitly mean a “function in / of $x$”. However, noone speaks of $f$ as a “function in / of $x$” (at least I haven’t heard that). Thereby, both usages may live together in peace as long as one exercises caution. $\endgroup$
    – k.stm
    Commented Nov 26, 2017 at 8:26
  • $\begingroup$ @k.stm: I agree in principle, but have personally found it almost impossible to exercise this caution in practice. For example, if one wishes to consistently and explicitly make the distinction while teaching calculus, one ends up constantly saying things like: "the function of $x$, $x^2$...". This was not necessary during the 200 years prior to 1910, as $f$ was not yet called a function then. $\endgroup$
    – Michael Bächtold
    Commented Nov 26, 2017 at 16:14
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    $\begingroup$ one could "dependent variable" as in books on ODEs. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Nov 26, 2017 at 21:06
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    $\begingroup$ @k.stm "noone speaks of $f$ as a "function of $x$"". Well, every calculus book writes $\frac{\partial f}{\partial x}$ when $f:\mathbb{R}^2\to \mathbb{R}$, which is basically another way of doing the same. The origin of this seems to be the article of Jacobi. $\endgroup$
    – Michael Bächtold
    Commented Nov 29, 2017 at 14:36
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The Fourier transform is defined in at least 3 different ways depending on which subject and school one comes from: $$ \hat{f}(\xi)=\int_{\mathbb{R}^n} f(x)e^{-2\pi ix\cdot\xi}dx $$ or $$ \hat{f}(\xi) = \int_{\mathbb{R}^n} f(x)e^{-ix\cdot\xi}dx $$ or $$ \hat{f}(\xi) = \frac{1}{(2\pi)^{\frac{n}{2}}}\int_{\mathbb{R}^n} f(x)e^{-ix\cdot\xi}dx $$ The second definition is no longer an unitary operator on $L^2(\mathbb{R}^n)$ and loses the usual symmetry with the inverse, both of which are however rectified again by the third definition.

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    $\begingroup$ Also, some definitions of the Fourier transform will change the complex exponential's sign. $\endgroup$
    – J.G.
    Commented Dec 3, 2017 at 22:52
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    $\begingroup$ For the second definition you may use $L^2$ space with the measure $(2\pi)^{-d} d\xi$. This is sometimes used in microlocal analysis to get rid of all the annoying powers of $2\pi$. $\endgroup$
    – mcd
    Commented Apr 28, 2018 at 11:12
  • $\begingroup$ I use the second in teaching (engineers) and the first in research. Including a factor in the definition to make the $L^2$ theory fluid is not good in a course, I think. And the $L^1$ theory will have the extra factor. $\endgroup$
    – Nicola Arcozzi
    Commented Jun 15, 2019 at 10:06
  • $\begingroup$ @NicolaArcozzi, re, what does it mean to make the $L^2$ theory fluid? $\endgroup$
    – LSpice
    Commented Sep 12, 2023 at 2:54
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How about "algebra"? Usually an algebra over a field is assumed to be associative by default, but sometimes it is not.

Not to mention the various category-theory uses of "algebra" (over a monad, over an operad, for a Lawvere theory, for an endofunctor...).

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    $\begingroup$ The category-theory sense of "algebra" is related to the one in Universal Algebra: An algebraic signature is a set of function symbols, each with a non-negative arity. An algebra (over a signature) is a any non-empty set together with n-ary functions for the n-ary function symbols. (These algebras are occasionally called universal algebras or algebraic structures.) $\endgroup$
    – user12283
    Commented Nov 23, 2017 at 13:48
  • $\begingroup$ @HansAdler Still, I think the average mathematician might be a least a bit confused if I claim that compact Hausdorff spaces are examples of "algebras". Also sometimes one has to be very precise, e.g. to not confuse algebras over a monad and algebras for the corresponding endofunctor. $\endgroup$
    – Stefan Perko
    Commented Nov 24, 2017 at 16:18
  • $\begingroup$ Not sure how you read it, but I just wanted to expand on your second paragraph by mentioning a definition of 'algebra' that most mathematicians find easier to relate to than category-theoretic notions of algebra, but which is still vastly more general than any of the classical notions of an algebra. - Btw, I wasn't aware that Hausdorff spaces are examples of the category theoretic notion of algebra. Is that by cheating (working in a category of Hausdorff spaces and with an empty signature)? $\endgroup$
    – user12283
    Commented Nov 28, 2017 at 15:42
  • $\begingroup$ @HansAdler Ah, I see. Yes, thank you. Feel free to edit my answer as well. - Well, I guess this the problem with the word "algebra". Noone knows what it means. I'm kidding. I was referring to the fact that compact Hausdorff spaces are monadic over $\mathsf{Set}$. This is "Handbook of categorical algebra 2" Proposition 4.6.6. $\endgroup$
    – Stefan Perko
    Commented Nov 28, 2017 at 20:16
  • $\begingroup$ @HansAdler I guess the choice of the word "algebra" in universal algebra is the main reason this is not widely known. (Calling the field universal algebra is not a problem, but calling its structures "algebras" is dire) $\endgroup$
    – YCor
    Commented Jan 2, 2019 at 10:11
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What is a "set"? ZFC has one answer; ETCS has another; Bishop had another; HoTT has yet another.

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  • $\begingroup$ Are they really part of the same field? $\endgroup$
    – Michael Greinecker
    Commented Nov 22, 2017 at 22:10
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    $\begingroup$ @MichaelGreinecker Yes and no. No in that people who study ZF-sets for their own sake are a distinct community from people who study ETCS-sets and people who study HoTT-sets. But yes in that all of them have been proposed as a precise meaning of the word "set" as it is used foundationally in the rest of mathematics. $\endgroup$
    – Mike Shulman
    Commented Nov 22, 2017 at 22:18
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    $\begingroup$ Maybe the biggest conceptual difference is between ZFC on the one hand, and ETCS and HoTT on the other (I don't know where Bishop lies). In ZFC, a "set" is (conceptually) a subset of the universal class; set membership is a predicate and one set can be a subset of another. In HoTT, set membership is not a predicate and an element of a set cannot also belong to another set. But there is a separate concept, "subset of the set $S$", where membership is a predicate (over $S$), and one subset-of-$S$ can be a sub-(subset-of-$S$) of another subset-of-$S$. $\endgroup$
    – Sophie Swett
    Commented Nov 22, 2017 at 23:04
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    $\begingroup$ The reason for this is that HoTT (and I think also ETCS, but I'm not sure) are type theories, in which "sets" are a particular kind of type, and all types are disjoint from each other. The word "set" does not refer to a subset of a type. ZFC, on the other hand, only has one "type", which is the universal class; and the word "set" does refer to a subset of this "type". $\endgroup$
    – Sophie Swett
    Commented Nov 22, 2017 at 23:07
  • $\begingroup$ @TannerSwett ETCS is not technically a type theory, but it shares many features of type theories. $\endgroup$
    – Mike Shulman
    Commented Nov 23, 2017 at 0:04
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To many authors a "category" is necessarily locally small, but to others it need not be.

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    $\begingroup$ Mac Lane understands small to mean “within some fixed Grothendieck universe $U$” – is this nowadays completely out of fashion or do usages of this term here, too? $\endgroup$
    – k.stm
    Commented Nov 26, 2017 at 8:33
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    $\begingroup$ Sometimes "small" means that; other times it means "a set" (as opposed to a proper class). But I wouldn't consider that two different meanings for the same word, rather two different ways to formalize the same meaning. (-: $\endgroup$
    – Mike Shulman
    Commented Nov 26, 2017 at 12:18
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Some authors formulate separability axioms T3/T4 as normality/regularity plus T1. Some other authors do not require T1, on the contrast, they define normality/regularity as T3/T4 plus T1.

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  • $\begingroup$ Really? I've seen "regular" used for what others called "regular Hausdorff", but I don't recall anyone using $T_3$ without also meaning $T_2$. Do you have any references? $\endgroup$
    – tomasz
    Commented Nov 27, 2017 at 17:18
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    $\begingroup$ I quote wikipedia: "Although the definitions presented here for "regular" and "T3" are not uncommon, there is significant variation in the literature: some authors switch the definitions of "regular" and "T3" as they are used here, or use both terms interchangeably." They do not provide a reference in this place, however. $\endgroup$
    – Fedor Petrov
    Commented Nov 27, 2017 at 19:55
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    $\begingroup$ Some authors (e.g., Cullen 1968, pp. 113 and 118) interchange axiom T_3 and regularity, and axiom T_4 and normality (from mathworld.wolfram.com/SeparationAxioms.html) $\endgroup$
    – Fedor Petrov
    Commented Nov 28, 2017 at 19:35
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Simply connected : for some authors, such a space is necessarily path-connected, for others, not.

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    $\begingroup$ For that matter, is the empty space connected? $\endgroup$
    – Mike Shulman
    Commented Nov 22, 2017 at 22:16
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    $\begingroup$ The empty space should not be seen as connected, for the same reason as units in a ring are not primes: you want the decomposition into connected components to be unique up to permutation. I'm sure others will disagree, though, so it's another good example :) $\endgroup$
    – Lukas Lewark
    Commented Nov 23, 2017 at 5:41
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    $\begingroup$ @LukasLewark I agree ... connected components are seen as "atoms" and therefore empty space should not be considered connected although this does not match with the formal definition. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Nov 23, 2017 at 10:50
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    $\begingroup$ This mini-debate is exactly the simply-connected+(conntected or not) debate one dimension down! $\endgroup$
    – Jeff Strom
    Commented Nov 23, 2017 at 14:49
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    $\begingroup$ The question about connectedness of the empty set is the question of whether the set of components of the empty set is $\emptyset$ or $\{\emptyset\}$. (Fortunately, according to my Discrete Mathematics students, there is no difference between these two sets.) $\endgroup$
    – LSpice
    Commented May 25, 2019 at 16:45
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"Topos" sometimes means "elementary topos" and sometimes "Grothendieck topos".

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    $\begingroup$ There is a correlation between the intended meaning and the pluralisation. Algebraic geometers usually study Grothendieck topoi, and logicians typically study elementary toposes. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Oct 16, 2018 at 18:04
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A tree can be a very different thing in different parts of mathematics. It might be a certain kind of acyclic graph; or a partial order such that the predecessors of every node are linearly ordered; or a partial order where the predecessors are well-ordered; or either of these, except with successors instead of predecessors. In some parts of mathematics, trees are presumed finite, while in others, they are nontrivial only when infinite.

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    $\begingroup$ For that matter, so can a "graph". $\endgroup$
    – Mike Shulman
    Commented Nov 22, 2017 at 21:59
  • $\begingroup$ ... and a field $\endgroup$
    – Fedor Petrov
    Commented Nov 22, 2017 at 22:00
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    $\begingroup$ @FedorPetrov If you mean "commutative division ring" versus "function on a space", then I don't think that's an example for this question, as noted in the comments comparing it to the other question about overloaded words. But the various meanings of "graph" and "tree" are very closely related, yet distinct. $\endgroup$
    – Mike Shulman
    Commented Nov 22, 2017 at 22:04
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    $\begingroup$ I interpreted Mike's remark to refer to the fact that graphs have different conceptions even in graph theory, depending on whether self-edges or parallel edges are allowed. $\endgroup$
    – Joel David Hamkins
    Commented Nov 22, 2017 at 22:11
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    $\begingroup$ If we were to go along this road, I'd suggest to look for all the various definitions of "normal" in different branches of mathematics ... $\endgroup$
    – Hagen von Eitzen
    Commented Nov 23, 2017 at 12:19
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I would say the word "kernel" is probably among the most overloaded terms in mathematics. You've got kernels of linear operators, convolutional kernels, distribution kernels, Markov kernels, and Reproducing Hilbert Space kernels. All of these notions are related but are, strictly speaking, distinct objects. Thus, the kernel of a linear operator is a subspace, while an RKHS kernel is a positive definite map from some set times itself to the reals.

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    $\begingroup$ And in graph theory, it refers to the result of removing all vertices of degree $\leq 1$ from a graph followed by 'de-subdividing' paths containing degree-2 vertices. $\endgroup$
    – Adam P. Goucher
    Commented Nov 22, 2017 at 21:39
  • $\begingroup$ Oh wow, that's quite distinct from the others! $\endgroup$
    – Aryeh Kontorovich
    Commented Nov 22, 2017 at 21:42
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    $\begingroup$ I don't think this is what the question was asking for; see the comment comparing it to the other question about overloaded words. $\endgroup$
    – Mike Shulman
    Commented Nov 22, 2017 at 22:06
  • $\begingroup$ Oh I see. Is there a way to migrate answers? $\endgroup$
    – Aryeh Kontorovich
    Commented Nov 22, 2017 at 22:06
  • $\begingroup$ I don't think so, but the other question is closed anyway. $\endgroup$
    – Mike Shulman
    Commented Nov 22, 2017 at 22:11
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Topology: Some authors use the term neighborhood while other use open neighborhood instead.

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    $\begingroup$ This is worse than you indicate. "Neighborhood" is defined in two different ways. J.L.Kelley (General Topology) defines neighborhood of $x$ as an open set containing $x$, then asserts that analysts and Bourbaki use "neighborhood" to be any set containing an open set containing $x$. (References are in my math.SE answer here.) $\endgroup$
    – Eric Towers
    Commented Nov 24, 2017 at 6:56
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    $\begingroup$ Most of the time people use neighbourhoods where it does not matter if it is open or not and the end result is the same. Problems begin if one insists on taking, say, complements or unions of neighbourhoods, in which case it is better to simply speak of open sets instead, or introduces a connectedness hypothesis in the definition of neighbourhood... $\endgroup$
    – M.G.
    Commented Nov 24, 2017 at 13:15
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    $\begingroup$ @July It matters very much when one wants to work with neighborhood filter. $\endgroup$
    – Michael Greinecker
    Commented Nov 27, 2017 at 8:40
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    $\begingroup$ @July The point is that filters are closed under supersets, so this pretty much requires the "French" definition. Or talking about the filter generated by open neighborhoods, but I think filters are a big part of what terminology is favored. $\endgroup$
    – Michael Greinecker
    Commented Nov 27, 2017 at 12:57
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    $\begingroup$ I think the language used in the theory of locally compact spaces (where we talk about compact neighbourhoods) and the theory of locally convex vector spaces (where we make statements like "the polar of an equicontinuous set is a neighbourhood of 0") would become very awkward if neighbourhood could only be used to mean "open neighbourhood". I can't think of an opposing situation in which it is beneficial to have "neighbourhood" mean "open neighbourhood". $\endgroup$
    – Robert Furber
    Commented Nov 27, 2017 at 23:01
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The definition of a Turing Machine is a great example, where multiplied all together there are at least hundreds of possible definitions.

  • Is the tape doubly infinite or singly infinite? (If singly infinite, what happens when you move left at the left at the end of the tape?)

  • Does the machine have special "accept" and "reject" states, just a "halt" state, or neither of these?

  • Is the input string written on a separate read-only tape? Is the output string written on a separate write-only tape?

  • Does the number of states include the accept and reject states?

  • Does the space usage include input and output? Does the running time include the step when the machine halts?

  • Is the input alphabet $\{0,1\}$, or an arbitrary set? May it contain only 1 symbol or even 0 symbols?

  • Is the tape alphabet just adding a blank symbol, or may it contain other convenient working symbols?

  • Are multiple tapes allowed? Is staying still instead of moving left or right allowed?

One usually spends a week in a theory of computation course fleshing out some of the equivalences between various definitions. But I don't know that anyone has ever done this in machine-checked or at least formally exhaustive detail, avoiding all pitfalls and edge cases. Rather, researchers rely heavily on the informal Church-Turing thesis and assume some kind of programming language or machine representation which suits their needs and is equivalent to all these definitions of Turing Machine at once.

Yet they are not at all equivalent for many purposes. (1) If worried about low-level running time (not just up-to-a-polynomial), doubly-infinite vs singly-infinite, how many symbols are allowed, multiple vs. single tapes, and whether accept/reject is part of the finite state machine all have an impact. (2) If worried about sub-linear complexity classes, we know that it is crucial that the input string be read-only. (3) Even if only worried about defining polynomial time computability, at least two symbols are required in the input alphabet. (4) To define the Busy Beaver function, as well as many other non-computable functions, requires a specific representation to be fixed. This one is a big "gotcha" that one always runs into when asking students (as an exercise) to compute the Busy Beaver function of $1$ or $2$.

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    $\begingroup$ These subtleties have a dramatic effect on computable analysis $\endgroup$
    – wlad
    Commented Aug 10, 2023 at 14:17
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Perhaps a prominent example is the definition of a smooth manifold. Some authors require the underlying locally Euclidean Hausdorff space to be 2nd countable, while others require it to be only paracompact. The latter is more general since it allows for uncountable number of connected components, although I'm not sure if that makes such a big difference in the everyday life of a differential geometer. But usually there is no room for confusion since most authors state at the beginning which convention they use.

EDIT: It turns out it actually does make a pretty big difference, see Stefan Waldmann's comment below.

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    $\begingroup$ It actually does make a huge difference: any set can be a zero-dimensional manifold if you only require paracompactness. In particular, nice theorems like: a bijective immersion is a diffeomorphism will fail in this context while they hold in the 2nd countable context. A simple consequence/example: a transitive Lie group action of $G$ on $M$ will not yield a homogeneous space structure $G/G_p$ for $M$ if you are allowed to topologize $G$ discretely... Very nasty. $\endgroup$
    – Stefan Waldmann
    Commented Nov 23, 2017 at 6:31
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    $\begingroup$ @July Consider $\mathbb{R}^0$. $\endgroup$
    – Robert Furber
    Commented Nov 23, 2017 at 13:36
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    $\begingroup$ @RobertFurber: right, thanks, I forgot that counts as Euclidean, I feel so silly now. So, ironically, my other comment about excluding $0$ from the naturals came back to hunt me :D $\endgroup$
    – M.G.
    Commented Nov 23, 2017 at 13:40
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    $\begingroup$ When I first learned of n-manifolds in the early 1960s, they were required only to be locally homeomorphic to n-dimensional euclidean space. Then people started adding additional conditions like Hausdorff, second-countable or paracompact — to the very definition of the term. Then there are also the topological, piecewise-linear, C^r (1 ≤ r ≤ ∞), and real-analytic (C^𝜔), not to mention complex manifolds. So many definitions make it a very good idea for anyone using the term "manifold" to state exactly what the assumptions are. $\endgroup$
    – Daniel Asimov
    Commented Aug 10, 2023 at 20:01
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    $\begingroup$ The so-called Prüfer manifold is a real-analytic surface that is not second countable. It is extremely elegant, and in fact its complexification is a 2-dimensional complex manifold that is also not second countable — which thanks to Radó's theorem every Riemann surface must in fact be. I would hate to exclude the Prüfer manifold on the grounds that it is "not a real manifold". $\endgroup$
    – Daniel Asimov
    Commented Aug 10, 2023 at 20:17
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Two examples from point-set topology:

  • a topological space $(X,\tau)$ is locally compact if "any point has a compact neighborhood" or if "any point has a local basis of compact neighborhoods". The two are equivalent if the space is Hausdorff, but not in general!
  • Even more subtle: a topological space $(X,\tau)$ is locally connected at $x$ if "$x$ has a local basis of connected neighborhoods" or if "$x$ has a local basis of open connected neighborhoods". The two are equivalent if we ask them for all points $x\in X$, but not in general!
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    $\begingroup$ The second point is actually frustrating in practice. Most authors use connected im-kleinen for your first definition of locally connected, but I've also seen weakly locally connected. $\endgroup$
    – Forever Mozart
    Commented Jan 2, 2019 at 7:23
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Antisymmetrization and symmetrization of tensors. Should we divide it by $(n!)$ ? This affects the relations between tensor and (anti-)symmetric algebra, the theory over $\mathbb Q$ and $\mathbb Z$ and is generally a mess. A special case: is a quadratic form over $\mathbb Z$ represented by a polynomial with integer coefficients or by a symmetric matrix with integer elements?

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    $\begingroup$ It also affects whether you can symmetrise or antisymmetrise in positive characteristic …. $\endgroup$
    – LSpice
    Commented May 25, 2019 at 16:49
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$(a,b)$

Is that a coordinate pair representing a point in the plane? or,

The open interval from $a$ to $b$? or

The greatest (highest) common factor (divisor) of $a$ and $b$? or

The ideal generated by $a$ and $b$? or

The inner product of $a$ and $b$? or...

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    $\begingroup$ This is surely a case of notation, rather than a definition $\endgroup$
    – Yemon Choi
    Commented Nov 26, 2017 at 20:30
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    $\begingroup$ This was my spontaneous answer to the part of the question saying that "different places in the literature the same name is used for two different mathematical objects." Plus, I am thinking of a notation as a written form of a name. $\endgroup$
    – Amir Asghari
    Commented Nov 26, 2017 at 21:31
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    $\begingroup$ This is why I always use $\langle a,b\rangle$ for points in the plane. I've even used $\langle a,b,c,...,h,...\rangle$ to indicate points in the hilbert cube! $\endgroup$
    – Forever Mozart
    Commented Jan 2, 2019 at 7:18
  • $\begingroup$ @ForeverMozart, re, what is the 27th component of that point in the Hilbert cube? $\endgroup$
    – LSpice
    Commented Sep 12, 2023 at 2:52
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For some strange reason, algebraic order theory choose "lattice theory" as its name, and "lattice" as its central object of study. But of course, for everybody else "lattice" still means what it always meant, and what it also means in plain English.

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    $\begingroup$ Even for the "classical" lattices, there are many different definitions depending on the book or paper you look at, and some of them only coincide for the special case of a full, free $\mathbb{Z}$-module in a $\mathbb{Q}$-vector-space... $\endgroup$
    – Dirk
    Commented Nov 23, 2017 at 10:36
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    $\begingroup$ The term lattice is also used in functional analysis, probability theory, computer science, economics,... Everybody else might be a minority nowadays. $\endgroup$
    – Michael Greinecker
    Commented Nov 23, 2017 at 13:10
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    $\begingroup$ This would be a better answer if you spelled out what lattice "always meant". In any case, this seems to me to be more an example of a common word being overloaded. Certainly the name "lattice" for the algebraic structure seems very reasonable once you see a picture of a Hasse diagram. $\endgroup$
    – Alex Kruckman
    Commented Nov 29, 2017 at 16:27
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    $\begingroup$ (cont'd) Also, one could speak about a Lie(-theoretic) lattice, or lattice subgroup, when referring to the discrete subgroup of Lie group case. $\endgroup$
    – Qfwfq
    Commented Apr 28, 2018 at 13:38
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    $\begingroup$ This one has confused the Library of Congress. For example, Bremner's book on lattice basis reduction has an LC call number of QA171.5 .B74 2012 which lumps it together with Birkhoff's book on lattice theory. Bremner's book should really be QA 9.58 something. $\endgroup$
    – Timothy Chow
    Commented Sep 1, 2019 at 16:54
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The term 'Macdonald polynomial' might refer to:

  • The symmetric function $P_\lambda(x;q,t)$ with coefficients being rational in $q,t$.
  • The symmetric function $J_\lambda(x;q,t)$, being $P_\lambda(x;q,t)$ multiplied by a normalization factor.
  • The one of the non-symmetric polynomial $E_\alpha(x;q,t)$ (Haglund vs. Marshall normalization/convention).
  • The non-symmetric polynomial $E_\lambda(x;q,t)$ indexed by weights in a root lattice, and there is a formula over alcove walks.

  • The modified Macdonald polynomial $\tilde{H}_\lambda(x;q,t)$, which is the most 'combinatorial' version, which is the bigraded Frobenius character of a certain $S_n$-modules indexed by $\lambda$. Combinatorial formula due to Haglund.

  • There are also non-homogeneous Macdonald polynomials...

This specializes to confusion regarding Hall-Littlewood polynomials (the 'standard' $P_\lambda$, the 'dual' $Q'_\lambda$ or the modified, $\tilde{H}_\lambda$).

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An extension of a group $A$ by a group $B$ can be either a group $G$ with a normal subgroup isomorphic to $B$ with $G/B$ isomorphic to $A$ or a group $G$ with a normal subgroup isomorphic to $A$ with $G/A$ isomorphic to $B$. (See, for example, https://en.wikipedia.org/wiki/Group_extension especially section 1.2.3.)

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    $\begingroup$ This is why you always just show the damn exact sequence. $\endgroup$
    – darij grinberg
    Commented Nov 26, 2017 at 19:41
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    $\begingroup$ Even more confusion can be had when speaking of a $P$-by-$Q$ group (to pick a random example, "solvable-by-finite"). $\endgroup$
    – LSpice
    Commented Dec 1, 2017 at 15:53
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In a real vector space $E$, some people use cone to mean a subset $C \subseteq E$ closed under multiplication by positive reals. If it is additionally closed under addition, they call it a convex cone, and if $C \cap -C = \{0\}$, they say it is a proper cone, or that $C$ is proper.

A second group of people has no name for the first thing, but calls the second thing just a cone, also using proper cone for the third thing.

A third group of people also never use the first thing, but call the second thing a wedge and the third thing a cone. So instead of saying "we prove that this cone is proper", they say "we prove that this wedge is a cone".

Since the preorder defined by a convex cone is a partial order iff that cone is proper, I am inclined to put myself in the third group of people and use the short word cone for the most useful notion.

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  • $\begingroup$ I would have said multiplication by nonnegative reals. $\endgroup$
    – Daniel Asimov
    Commented Aug 10, 2023 at 20:07
  • $\begingroup$ @DanielAsimov See my comment to another answer to this same question. "Positive reals" means reals $\geq 0$ in functional analysis, those $> 0$ are "strictly positive reals". $\endgroup$
    – Robert Furber
    Commented Nov 8, 2023 at 11:04
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    $\begingroup$ Wait until the fact 0 is sometimes considered a positive integer in France is brought up. But more seriously, this is not worth arguing over. (I have deleted some unproductive comments, but left the first two to highlight the ambiguity and the clarification) $\endgroup$
    – David Roberts ♦
    Commented Nov 10, 2023 at 2:18
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Speaking of functional analysis: While algebraists seem (at least as far as I can tell) to finally accept that things called ring and in particular things called algebra really should have units, people in functional analysis seem adamantly opposed to this, especially those who work with $C^\ast$-"algebras". I can sort of see why they do it, but I don't find their reasons particularly compelling; they rarely seem to talk about actual honest to god $C^\ast$-algebras-without-unit. Most of the time an "algebra" is really better viewed as an ideal in some proper algebra. All the proofs I have seen work the same way: First step is always to go to the unital case by embedding the "algebra" in question into an unital algebra.

As an algebraist myself, talking to a friend of mine who lives in $C^\ast$-land sometimes is a hassle because of this.

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    $\begingroup$ What? We talk about nonunital C*-algebras all the time. "All the proofs I have seen" --- well, then you're not very familiar with the subject. $\endgroup$
    – Nik Weaver
    Commented Nov 22, 2017 at 21:51
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    $\begingroup$ Adding a unit is certainly a technique that gets used, but there's a famous quote by Gert Pedersen about how "in the old days" we used to always start by adding a unit, but now we start by tensoring with the compacts and getting rid of the unit. It's an overstatement, but tensoring with the compacts is certainly also a common technique. $\endgroup$
    – Nik Weaver
    Commented Nov 23, 2017 at 1:27
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    $\begingroup$ Upvoted because this is a really important case of conflicting definitions, but I wish you’d mentioned the related question of whether ring homomorphism means unital and left out the polemics about which was right. $\endgroup$
    – Noah Snyder
    Commented Nov 23, 2017 at 15:42
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    $\begingroup$ Well, these are more nuanced and more interesting comments. But would you also tell topologists that locally compact Hausdorff spaces should not be considered as "proper" topological spaces, but rather as open subsets of compact Hausdorff spaces? $\endgroup$
    – Nik Weaver
    Commented Nov 23, 2017 at 18:38
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    $\begingroup$ Also, you didn't mention K-theory, which is a good example of an area where you really want nonunital C*-algebras. You nonunitally embed $A$ in $M_2(A)$ in $M_3(A)$ etc. as upper-left-corners, then you take the nonunital direct limit to get the nonunital algebra $A \otimes K(H)$, and it is the projections living in here that yield the $K_0$ group of $A$. $\endgroup$
    – Nik Weaver
    Commented Nov 23, 2017 at 18:40
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Variety has a number of slightly different definitions. Apparently some authors use reduced of finite type over a field, whereas I would want separated to rule out the line with two origins. Most people use scheme-theoretic language to solve this problem.

Conway-Sloane jokes the discriminant of a quadratic form is a function of the form and the author. They are not wrong. That same subject has two distinct definitions of integral, which are basically the same (multiply by $2$) but which are a pain to keep track of. The Hasse-Witt invariant has two different definitions as well.

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  • $\begingroup$ Re: quadratic forms: What's odd to me is Gauss went "twos in" in Disquisitiones and "everyone else" went "twos out". This has led to pairs of theorems that address this both ways, e.g. the 15 and 290 theorems. $\endgroup$
    – Eric Towers
    Commented Nov 24, 2017 at 6:45
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