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The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) $\sin z$ is a bounded function;
(iii) $\sin z$ is defined and analytic everywhere on $\mathbb{C}$;
(iv) $\sin z$ is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of $\sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset $U$ of $\mathbb{R}$ must be the whole of $\mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

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    $\begingroup$ I have to say this is proving to be one of the more useful CW big-list questions on the site... $\endgroup$ – Qiaochu Yuan May 6 '10 at 0:55
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    $\begingroup$ The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. $\endgroup$ – Unknown May 22 '10 at 9:04
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    $\begingroup$ wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. $\endgroup$ – Suvrit Sep 20 '10 at 12:39
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    $\begingroup$ It's a thought -- I might consider it. $\endgroup$ – gowers Oct 4 '10 at 20:13
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    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$ – user9072 Oct 8 '11 at 14:27

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Maybe I am a bit late to the party, but here is something that I falsely believed in for a while:

False Belief: If $f$ is a continuously differentiable function with a horizontal asymptote, i.e. $\lim_{x\to +\infty}f(x)=L$, then $f'(x)\to 0$ as $x$ goes to infinity.

Of course, this is incorrect even for smooth functions; take $g$ to be a smooth $L^1$ function with oscillatory behavior so that $\lim_{x\to +\infty}g(x)$ does not exist and then set $f(x)=\int_{0}^xg(s)ds$.

I came to this misconception during an ODE course I took in my undergrad. In the study of the asymptotic behavior of solutions of an autonomous equation $$ \frac{dy}{dx}=F(y) $$ one argues that if a solution $y=y(x)$ tends to a limit $L$ as $x\to +\infty$, that limit must be a zero of the vector field. We then proceeds with determining which zeros of $F$ are asymptotically stable to find $L$. The point is that here not only $\lim_{x\to +\infty}y$, but also $\lim_{x\to +\infty}\frac{dy}{dx}$ exists due to the ODE. Hence:

Correct Statement: If $f$ is a continuously differentiable function with a horizontal asymptote, and if $f'$ admits a limit as $x$ goes to infinity, that limit must be zero.

This readily follows from the mean value theorem.

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    $\begingroup$ A related old answer. I have found that people usually need some prompting to think of an $L^1$ function that does not converge to $0$ at infinity, even though all one needs is a sequence of spikes of increasing height, but sufficiently rapidly decreasing width so that the sum of the areas under them converges. $\endgroup$ – Robert Furber Mar 3 '20 at 17:02
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It is a common mistake to believe that epimorphisms are either identical to surjections or that they are a better concept. Unfortunately this is rarely the case; epimorphisms can be very mysterious and have unexpected behavior

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    $\begingroup$ Can you give a concrete example why it is a "mistake"? $\endgroup$ – Willie Wong Apr 25 '20 at 13:15
  • $\begingroup$ If you mean set theoretic surjective then in every concrete category a morphism is surjective then the morphism is epi and in all the category that I know epi are something very reasonable for example in Hausdorff spaces epi are exactly the map with dense image ... $\endgroup$ – Anonyme Sep 16 '20 at 13:00
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    $\begingroup$ I think the best example where this constitutes a mistake is the category of rings. For example the inclusion $\mathbb{Z}\to \mathbb{Q}$ is an epimorphism. $\endgroup$ – Louis Hainaut Oct 16 '20 at 11:39
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I already thought that the following two sets of matrices are one. $$M(\color{blue}{\Bbb R},2n)\qquad \text{and}\qquad M(\color{red}{\Bbb C},n).$$

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  • $\begingroup$ ... matrices, presumably? $\endgroup$ – AlexArvanitakis Nov 5 '20 at 4:17
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Maybe that's a cute one:

If $f$ is a continuous real function, then $f^{-1}(x)$ is at most countable unless $f$ is constant on some interval.

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    $\begingroup$ How would you stumble upon this belief? The most basic example of continuous functions (constant ones) fail it, so it seems like an easy belief to correct. $\endgroup$ – Mark Nov 14 '20 at 21:00
  • $\begingroup$ Well, thanks, clarified what was left as obvious. $\endgroup$ – A.DellaCorte Nov 14 '20 at 21:04
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    $\begingroup$ Hmm, with the clarification added, really not obvious. What is an example of a counterexample? $\endgroup$ – JoshuaZ Nov 14 '20 at 23:21
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    $\begingroup$ Any component of a Peano curve does the job. $\endgroup$ – A.DellaCorte Nov 14 '20 at 23:59
  • $\begingroup$ You don't need that fancy Peano curve. Every closed set in a metric space is the zero set of a real valued function. $\endgroup$ – Yai0Phah Feb 8 at 22:43
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This is more sort of a convention issue than an outright false belief (connected to the usual $\emptyset$ vs $\{\emptyset\}$ stuff), but I find it funny. I guess a fair share of mathematicians believe that: \begin{equation} \bigcap\emptyset=\emptyset\label{eq}\end{equation} while retaining the standard definition for intersection: $$\bigcap S:=\{x\ \text{such that}\ \forall Y(Y\in S\implies x\in Y)\}$$ according to which in fact: $$\bigcap\emptyset=V$$ where $V$ is the universal class. The condition in round brackets is of course vacuously true. So in a way - this is what I find funny - the former is the worst possible tentative solution of an equation ever.

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Here is a false belief I've been carrying in my head for 5 years and only woken up from recently. When one defines symmetric and anti-symmetric tensors on a vector space $V$, one usually starts with an action of the symmetric group on tensors, which is usually defined as:

$$ S_n \curvearrowright V^{\otimes n}, \quad \sigma (v_1 \otimes \ldots \otimes v_n) = v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}. $$

Now, if $\tau \in S_n$ is another permutation, we have

$$ \tau(\sigma (v_1 \otimes \ldots \otimes v_n)) = \tau(v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}) = v_{\tau(\sigma(1))} \otimes \ldots \otimes v_{\tau(\sigma(n))}, $$

right? Wrong! The action above is a right action, so

$$ \tau(v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}) = v_{\sigma(\tau(1))} \otimes \ldots \otimes v_{\sigma(\tau(n))} = (\sigma \cdot \tau)(v_1 \otimes \ldots \otimes v_n). $$

The reason for that is that the defining formula for the action doesn't directly tell you where $v_i$ goes. Rather, it shows who comes to the $i$-th place.

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    $\begingroup$ This is something I have reteach myself every time I teach it to others. Happily I remember that I need to do so. $\endgroup$ – Simon Wadsley Feb 7 at 19:24
  • $\begingroup$ Probably I will now demonstrate another false belief, but how is this possible? A $\tau$ is presented with an $v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)}$ to act on. It is not told what $\sigma$ was, how can it guess? In other words, this $v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)}$ is just some (rank 1) tensor, and $\tau$ is just supposed to permute its entries, without knowing anything about any previous actions on it, no? $\endgroup$ – მამუკა ჯიბლაძე Feb 9 at 6:05
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    $\begingroup$ @მამუკა ჯიბლაძე, one's mind gets tricked because in the original tensor, for each $v_i$, its subscript number coincides with its place within the tensor. And the action of $S_n$ is unfortunately formulated in terms of the subscripts. If you look at the definition of the action, it actually says: $\sigma$ takes a decomposable tensor and sends its $i$-th factor to the $\sigma^{-1}(i)$-th place. This inverse explains why the action is actually from the right. $\endgroup$ – Ivan Solonenko Feb 9 at 8:15
  • $\begingroup$ Thank you! I overlooked your statement that the action is from the right, and was thinking about the left action that permutes the vectors according to $\sigma$. It is true that for different spaces $V_1$, ..., $V_n$ the simplest things available are canonical isomorphisms $V_1\otimes...\otimes V_n\to V_{\sigma(1)}\otimes...\otimes V_{\sigma(n)}$ and this gives that right action when the spaces coincide. When they do coincide there also is the left action that I had in mind but maybe it can also be defined abstractly? Say, in any symmetric monoidal category? $\endgroup$ – მამუკა ჯიბლაძე Feb 9 at 9:45
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    $\begingroup$ The same already happens with the permutation action of $S_n$ on $k^n$: the action $\sigma \cdot (x_1,\ldots,x_n) = (x_{\sigma(1)},\ldots,x_{\sigma(n)})$ is a right action. The better definition is $(\sigma x)_i = x_{\sigma(i)}$; the confusion comes from viewing coordinates as maps into versus maps out of your space. $\endgroup$ – R. van Dobben de Bruyn Feb 10 at 1:38
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I don't think I've seen it in here:

Every vector space has a non-trivial dual space ($L^p$ for $0 < p < 1$ was a counter-example only mentioned during one of the classes in measure theory)

And of course there's the common false belief of people outside of mathematics that "mathematicians work with numbers and formulae all day long" :)

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    $\begingroup$ Well, it is true that every vector space has a dual space, even $L^{1/2}$... and it is even true that every topological vector space has a continuous dual space... What you mean is that it is not true that every topological vector space has a non-trivial continuous dual space (or, that the continuous dual of a topological vector space does not necessarily separate points) $\endgroup$ – Mariano Suárez-Álvarez Jul 7 '10 at 18:54
  • $\begingroup$ You are indeed correct. I'll do better not to dismiss the trivial case the next time. $\endgroup$ – Asaf Karagila Jul 7 '10 at 19:31
  • $\begingroup$ The existence of a nonzero functional on a locally convex space is guarenteed by the Hahn-Banach theorem. $p$-Banach spaces are in general not locally convex if $p<1$. $\endgroup$ – Yai0Phah Feb 8 at 22:45
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These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.

Another mistake is if one i asked to solve this equation, $ \displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.

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  • $\begingroup$ What "people"? Non-mathematicians? $\endgroup$ – Todd Trimble May 4 '11 at 0:03
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    $\begingroup$ @Todd: No i was talking of high school students. $\endgroup$ – crskhr May 4 '11 at 4:08
  • $\begingroup$ @S.C.:if squarring both sides will not give the solution then how can second problem be solved? $\endgroup$ – Styles Oct 25 '17 at 16:08
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Hopefully this isn't a repeat answer. False belief: a matrix is positive definite if its determinant is positive.

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    $\begingroup$ Is this really a common(!) false belief? $\endgroup$ – Martin Brandenburg Oct 3 '11 at 7:23
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In ${\mathbb F}_p^\times$, the non-squares are the opposite of the squares. In other words, $a$ is square iff $-a$ is not a square.

This is a confusion with the facts that the kernel of $x\mapsto x^2$ is $\{1,-1\}$ and the subgroup of squares has index $2$.

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A few mistakes I remember:

  • The quotient groups $\frac{G}{N}$ and $\frac{H}{K}$ are isomorphic if $G \thicksim H$ and $N\thicksim K$.
  • A closed interval of a complete lattice is a complete sublattice.
  • Two homeomorphic topologies on a set are the same.
  • The set of all compatible uniformities of a topological group forms a complete lattice.
  • The trace of the identity matrix is 1.
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    $\begingroup$ A closed interval of a complete lattice does form a lattice that is complete, right? So that the mistake is that sups and infs in the interval (particularly the sup and inf over the empty set) are not necessarily computed as they would be in the ambient complete lattice; is that what you have in mind? $\endgroup$ – Todd Trimble Sep 6 '15 at 1:47
  • $\begingroup$ Yes‌‌‌‌‌‌‌‌‌‌‌‌. $\endgroup$ – Minimus Heximus Sep 6 '15 at 1:55
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Most people that study Riemannian geometry for their first time make the following assumption at some point: "Let $(e_1,\dots,e_n)$ be a local orthonormal frame of $TM$ such that all Lie brackets $[e_i,e_j]$ vanish..."

This one is not so common (maybe special to me), but here we go: "$\mathbb{RP}^\infty$ and $\mathbb{CP}^\infty$ are Eilenberg-Mac Lane spaces, so $\mathbb{HP}^\infty$ is one, too."

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    $\begingroup$ Although it's not an Eilenberg-MacLane space, $\mathbb{HP}^\infty$ is a classifying space (specifically $B(SU(2))$), just as $\mathbb{CP}^\infty = B(U(1))$ and $\mathbb{RP}^\infty = B(\mathbb{Z}/(2))$. $\endgroup$ – Robert Furber Mar 3 '20 at 15:55
  • $\begingroup$ @RobertFurber That's right. And because $BO(1)$ and $BU(1)$ are Eilenberg-Mac Lane spaces, you can immediately classify real and complex line bundles by a single cohomology class. But not quaternionic line bundles. $\endgroup$ – Sebastian Goette Mar 4 '20 at 19:46
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(1) All Lebesgue-null sets are countable, or are strongly measure zero. (2) The following,verbatim, was a Q in American Mathematical Monthly : " A student asserted that any uncountable real set has a closed uncountable subset. Is this true ?" .

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People seem to believe that conventional computation (for example, running a chaotic irreversible cellular automaton) can be as efficient as one wants simply with good engineering, but this is not the case. Landauer's principle states that erasing a bit of information always takes $\ln(2)\cdot k\cdot T$ energy where $k$ is Boltzmann's constant ($k=1.38065\cdot 10^{-23}$ Joules/Kelvin) and $T$ is the temperature. Landauer's principle is a consequence of the second law of thermodynamics since if Landauer's principle were violated, then entropy would decrease. Landauer's principle means that conventional irreversible computation always must take $\ln(2)\cdot k\cdot T$ energy per bit erased (and one can erase data just by running it through AND and OR gates, so every irreversible gate must take a minimum amount of energy by Landauer's principle). However, Landauer's principle does not apply to reversible computation since reversible computers are not allowed to erase data.

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    $\begingroup$ Okay, I see what you are aiming at with this last edit. The idea that ordinary computation can be made arbitrarily efficient is a reasonably common false belief about our physical world (and may even have a somewhat solid mathematical interpretation). I withdraw my initial objection. $\endgroup$ – S. Carnahan Aug 11 '17 at 14:04
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False belief: It is obvious how to prove that $\sin'=\cos$.

Not so much... if $\cos$ and $\sin$ are defined geometrically. You need to prove geometrically that $$\lim\limits_{x\to 0}\frac{\sin x}x=1$$ and a (non-circular) proof of that is not obvious (see here).

Personally I'm aware of that just today! (thanks to a remedial course given to my niece).

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    $\begingroup$ The string of comments below the Math.SE link shows that really pursuing this gets one down a rabbit hole of rigorous discussions of arc length and such. On the other hand, one can prove that if $S$ (think "sine") and $C$ (think "cosine") are continuous functions which satisfy the standard addition formulas and the Pythagorean theorem $C^2 + S^2 \equiv 1$, then $S'(0)$ exists and $S'(x) = S'(0)C(x)$. There is a whole family of sine-like functions $S_a: x \mapsto S(ax)$; adjusting the parameter $a$ so that $S_a^\prime(0) = 1$, you can define the standard sine to be this $S_a$: a neat finesse. $\endgroup$ – Todd Trimble Oct 8 '17 at 3:14
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    $\begingroup$ @ToddTrimble: Should we interpret your comment as for or against this false belief? Or else, is it just a neutral complement? For a full proof, is your way really easier? $\endgroup$ – Sebastien Palcoux Oct 10 '17 at 11:56
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    $\begingroup$ It's mainly in agreement with you, with a shade of neutral. I don't claim that my suggestion really makes it easier, but only that one can prove things rigorously without getting into considerations of arc length, with (theoretically anyway) no prerequisites past introductory differential calculus. Mainly it's based on convexity arguments; I have a write-up here: ncatlab.org/toddtrimble/published/…; see theorem 3.1 and the crucial lemma 3.4. The "finesse" is akin to how we adjust parameter $a$ in $f: x \mapsto a^x$ to $a = e$ to get $f'(0) = 1$. $\endgroup$ – Todd Trimble Oct 10 '17 at 13:22
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I once misunderstood the definition of monads, and thought that for a monad $(T,\eta,\mu)$, we have $T\eta_X = \eta_{TX}$ (or fmap return == return in Haskell). Of course this is not the case (in case of $T=$[], fmap return [1,2] is [[1],[2]], whereas return [1,2] is [[1,2]]).

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Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Constant_coefficients/

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:


There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\\ E_s\\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.


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    $\begingroup$ Dear Johannes, please reread my post. $\endgroup$ – Pierre-Yves Gaillard May 12 '10 at 15:18
  • $\begingroup$ Even a cursory examination of Nick Higham's book amazon.co.uk/Functions-Matrices-Computation-Nicholas-Higham/dp/… will show that both these opinions on the evaluation of matrix exponentials are hopelessly naive. $\endgroup$ – Robin Chapman May 15 '10 at 9:17
  • $\begingroup$ Dear Robin, Thanks for your answer. I don't have Nick Higham's book. I was wondering if you could be more precise. Your comment is very surprising to me: I thought I was stating a triviality. Here are two references en.wikipedia.org/wiki/Matrix_exponential#Alternative iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients Looking forward to hearing from you. $\endgroup$ – Pierre-Yves Gaillard May 15 '10 at 10:51
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    $\begingroup$ Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential. $\endgroup$ – Robin Chapman May 15 '10 at 14:07
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    $\begingroup$ I don't think the OP wants examples of normative statements. As I read it, the question is about conceptual errors regarding non-normative mathematical statements. $\endgroup$ – Qiaochu Yuan May 17 '10 at 6:19
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False belief: relativization is well-defined and the corresponding notation $C^A$ is unambiguous. Which is not quite true because $P=NP$ would not imply $P^A=NP^A$.

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    $\begingroup$ Maybe some more explanation would be useful. If decision problems and oracles are subsets of $\mathbb{N}$, and complexity classes are subsets of $P(\mathbb{N})$, then there is in general no such operation as relativization. I'm not sure how common of a false belief this is, but once I settled on my preference for the set point of view and saw what was going on here I lost some interest in the idea of relativization. $\endgroup$ – Dan Brumleve Jan 13 '18 at 5:16
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I used to think that the subset of even norm vectors in an integral lattice is a sub-lattice. This is true for the "classically integral" lattice defined by $<u,v> \in \mathbb{Z}$ for $u,v$ in the lattice because the even vectors is the kernel of the group homomorphism $v \rightarrow <v,v>$ mod 2. However this fails for the more general notion of "integer norm" lattice where we only require the quadratic form is integer valued (ie. the coefficients are all integral or that the off diagonal entries in the Gram matrix may be half integral eg $x^2+xy+2y^2$). For the hexagonal lattice $x^2+xy+y^2$ which is not classically integral, the even vectors is a sub-lattice but for a different reason that it is the lattice scaled by 2.

If $t = 3$ mod 4, the lattice $L_t$ with quadratic form $x^2+ty^2$ is classically integral. Its even sublattice $L_{t0}$ has quadratic form $4(x^2+xy+(t+1)y^2/4)$ which clearly equals 2$W_t$ where $W_t$ is the lattice with quadratic form $(x^2+xy+(t+1)y^2/4)$. If $t=7$ mod 8, the coefficients of the form is [odd,odd,even], the even vectors is not a subgroup since for example $[0,1]$ and $[1,1]$ has even norm but $[0,1]+[1,1]=[1,2]$ has odd norm. If $t$ is 3 mod 8, the form $(x^2+xy+(t+1)y^2/4)$ can only be even only if both $x,y$ are even since all coefficients are odd. So the even vectors in $W_t$ turn out to be $2W_t$. It is a sub-lattice and it is the subset of even vectors but it is index 4 in $W_t$. It is the 2-scaled sub-lattice.

If $v \in L_t$, $2v \in 2L_t \subset L_{t0}=2W_t$, so $L_t \subset W_t$. So the picture is $2W_t=L_{t0} \subset L_t \subset W_t$ with each containment is index 2.

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Let $M_1$ be a finitely generated module over a PID and let $M_2$ be a submodule.

We may pick $L_i$ and $T_i$ submodules of $M_i$ such that $L_i$ is free, $T_i$ is torsion, $M_i = L_i \oplus T_i$, $L_2\subseteq L_1$ and $T_2\subseteq T_1$.

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The following analogue of this result is clearly false:

Falsehood. If $M$ is a module over a commutative ring $R$, then $M^\vee = \operatorname{Hom}_R(M,R)$ is at least as big as $M$ (e.g. in terms of cardinality or rank).

For example, if $R$ is a domain and $M$ is torsion, then $M^\vee = 0$. But what's much more surprising is that the following is still false:

False belief: If $M$ is a torsion-free module over a principal ideal domain $R$ (even $R = \mathbf Z$), then $|M^\vee| \geq |M|$ and/or $\operatorname{rk}(M^\vee) \geq \operatorname{rk}(M)$.

(Even assuming $M$ has no divisble elements doesn't help.)

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    $\begingroup$ One might add that what is being overlooked here is that the "correct dualizer" is an injective cogenerator, rather than a projective generator such as $R$. What makes it work for vector spaces is that a field is an injective cogenerator over itself. By the way, what are the rings for which injective hull of $R$ works? It seems to always work for torsion frees but not for all - even over integers. Presumably $R$ must be local for that? $\endgroup$ – მამუკა ჯიბლაძე Feb 7 at 6:37
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Fans: (related to the one of polytopes written above) all convex cones are rational, i.e. one would expect that a line would eventually hit a point in the lattice. It is obviously not true, just take the one-dimensional cone generated by $(1,\sqrt{2})$. A similar one was thinking that if I rotate the cone a bit, I can always make it rational.

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    $\begingroup$ reminds me of the curious fact that some circles in the plane, too, have no points in $\mathbb Q^2$. (proven simply by cardinality!) $\endgroup$ – AndrewLMarshall Oct 4 '10 at 19:21
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False Belief: "The suspension spectrum map from spaces to (edit: symmetric) spectra preserves smash-products"

The facts that one denotes the smash product of spectra and the smash product of a space with a spectrum (levelwise) with the same $\wedge$ and tends to leave away the $\Sigma^\infty$ when one embeds a space into spectra are also not helpful in getting used to the harsh reality that the above is wrong.

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  • $\begingroup$ Yay! 100th answer! $\endgroup$ – Peter Arndt Oct 4 '10 at 21:33
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    $\begingroup$ I don't see that this qualifies as a false belief. In order for the question of whether it is true or false to even be meaningful, you have to first commit yourself to one of the many different notions of spectrum, not to mention smash product of spectra. $\endgroup$ – Tom Goodwillie Oct 5 '10 at 0:35
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    $\begingroup$ True. I meant symmetric spectra with the smash product coming from their description as modules over the symmetric sequence of spheres. $\endgroup$ – Peter Arndt Oct 5 '10 at 10:52
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If a matrix $A$ is self-adjoint/skew-self-adjoint with respect to a symmetric bilinear form, then it is diagonalizable.

True for matrices over $\mathbb{R}$, with respect to a positive definite inner product.

False over other fields. For example, over $\mathbb{C}$, $\left( \begin{smallmatrix} 1 & i \\ i & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & 1 & i \\ -1 & 0 & 0 \\ -i & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to the standard inner product.

False for other nondegenerate symmetric bilinear forms: $\left( \begin{smallmatrix} 1 & 1 \\ -1 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ -1 & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to $\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{smallmatrix} \right)$.

You can exponentiate the skew-self-adjoint matrices to get examples of matrices preserving a nondegenerate symmetric bilinear form, with Jordan blocks of the form $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$.

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    $\begingroup$ You seem to have a different definition of "the standard inner product on $\mathbb{C}^n$" than I do. I think that phrase normally refers to the familiar positive definite sesquilinear form, with respect to which self-adjoint matrices are indeed diagonalizable. $\endgroup$ – Mark Meckes Jan 28 '11 at 16:46
  • $\begingroup$ But that's not a bilinear form. And it has no generalization to other fields (what is it on $\overline{\mathbb{F}_p}$?). How can it be standard? :) I certainly agree that people should know that matrices which are self-adjoint with respect to the standard sesquilinear form are diagonalizable. $\endgroup$ – David E Speyer Jan 28 '11 at 17:51
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    $\begingroup$ Of course it's not bilinear -- an "inner product" on a complex vector space is defined to be sesquilinear, not bilinear -- I've spent a lot of time trying to get my linear algebra students to remember that. The failure of such a form to generalize to other fields is indeed sad, but I think the richness of Hilbert space theory helps to make up for that disappointment. :) $\endgroup$ – Mark Meckes Jan 28 '11 at 21:24
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If $a$ is a real zero of a cubic polynomial with rational coefficients then $a$ can be written as a combination of cube roots of rational numbers.

More generally if $a$ is a real zero of an irreducible polynomial with rational coefficients that is solvable by radicals then students expect the following:

  1. Any expression inside a radical evaluates to a real number
  2. Any sub-expression of the expression for $a$ evaluates to an algebraic number of order less than or equal to the order of $a$

Of course the problem is that from Cardan's solution to the cubic we can have negative rational numbers inside a square root. Let $c$ = $4*(-1 + \sqrt{-3})$.

$a$ = $\frac{\sqrt[3]{c}}{4} + \frac{1}{\sqrt[3]{c}}$

$f(x) = 4x^3 - 3x + \frac{1}{2}$.

So while $a$ is an algebraic number of degree three, it can not be written as combination of cube roots of rational numbers. Indeed, it is counter-intuitive that $\sqrt[3]{c}$ has degree 6 over the rational numbers yet we can use this number and simple arithmetic to produce an algebraic number of degree 3.

Also $a$ = $\sin(50^{\circ})$. For many values of $\theta$, $\sin \theta$ is a radical number. See also radical values for sine and cosine

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If every collection of disjoint open sets in a topological space is at most countable, then the space is separable

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Something I was sure about until earlier today:

Suppose $\kappa$ is an $\aleph$ number, then $AC_\kappa$ is equivalent to $W_\kappa$, namely the universe holds that the product of $\kappa$ many sets is non-empty if and only if every cardinality is either of size less than $\kappa$ or has a subset of cardinality $\kappa$.

In fact this is only true if you assume full $AC$, and $(\forall \kappa) AC_\kappa$ doesn't even imply $W_{\aleph_1}$, I was truly shocked.

Furthermore, $W_\kappa$ doesn't even imply $AC_\kappa$ in most cases.

The strongest psychological implication is that most people actually think of the well-ordering principle as a the "correct form" of choice, when it is actually Dependent Choice (limited to $\kappa$, or unbounded) which is the "proper" form, that is $DC_\kappa$ implies both $AC_\kappa$ and $W_\kappa$.

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    $\begingroup$ How common is this misconception? $\endgroup$ – Thierry Zell Apr 17 '11 at 3:08
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    $\begingroup$ @Thierry: For the past couple of weeks I spent a lot time considering models without choice, not only I held that misconception but not once anyone corrected me about it - grad students and professors alike. $\endgroup$ – Asaf Karagila Apr 17 '11 at 6:09
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Coordinates on a manifold do not have an immediate metric meaning. Until becoming familiar with differential geometry one tends to think they do. (Einstein wrote that he took seven years to free himself from this idea.)

For example, linear control theory is for the most part metric with variables in $R^n$. When moving away from linear control theory, variables are represented as coordinates on a manifold. Nevertheless, much of the literature tends to either abandon metric notions altogether, or to keep using an Euclidean metric though it is no longer very useful.

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The "conditional Vitali convergence theorem": Let $X_n$ be a uniformly integrable sequence of random variables with $X_n \to X$ almost surely, and $\mathcal{G}$ a sub-$\sigma$-field. Then $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ almost surely (FALSE).

I believed this one until I read Uniformly integrable sequence such that a.s. limit and conditional expectation do not commute. It is particularly startling because the conditional versions of the monotone convergence theorem, the dominated convergence theorem, and Fatou's lemma are all true!

What is true is that $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ in $L^1$, so you do have a subsequence converging almost surely.

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Here is a false belief I had. Let $f:X \to Y$ be a map of topological spaces having the property that for every finite CW complex $K$, the induced map $f_{\ast}:[K,X] \to [K,Y]$, on unpointed homotopy classes of maps, is a bijection. Then $f$ is a weak homotopy equivalence (that is, it induces isomorphisms on all homotopy groups relative to all basepoints). A counterexample is given by the stabilization map $B \Sigma_{\infty}\xrightarrow{+1} B \Sigma_{\infty}$, which is not an isomorphism on $\pi_1$.

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    $\begingroup$ Although the original intent of this question seems to have long since evaporated, I can't help asking: is this really a "common false belief"? $\endgroup$ – Yemon Choi Feb 17 '15 at 1:24
  • $\begingroup$ how about: if two CW complexes have all homotopy groups isomorphic, then they are homotopy equivalent? as i recall, you need those isomorphisms to be induced by a single continuous map. $\endgroup$ – roy smith Apr 22 '17 at 0:01
  • $\begingroup$ @roysmith Yes. You can even have two non weakly equivalent spaces having all Postnikov stages weakly equivalent $\endgroup$ – Ilan Barnea May 8 '17 at 10:47
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