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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... $\endgroup$ May 6, 2010 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, 2010 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, 2010 at 12:39
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    $\begingroup$ It's a thought -- I might consider it. $\endgroup$
    – gowers
    Oct 4, 2010 at 20:13
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    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$
    – user9072
    Oct 8, 2011 at 14:27

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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. $\endgroup$ Jan 28, 2011 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. $\endgroup$ Jan 28, 2011 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. :) $\endgroup$ Jan 28, 2011 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? $\endgroup$ Apr 17, 2011 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, 2011 at 6:09
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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? $\endgroup$ Oct 3, 2011 at 7:23
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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, 2015 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, 2017 at 0:01
  • $\begingroup$ @roysmith Yes. You can even have two non weakly equivalent spaces having all Postnikov stages weakly equivalent $\endgroup$ May 8, 2017 at 10:47
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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, 2017 at 14:04
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The GNS construction

Let $\phi$ be a state on a $C^*$-algebra $\mathcal{A}$, and put $N_{\phi}:=\{a\in\mathcal{A}\;|\;\phi(a^*a)=0\}$. Then $N_{\phi}$ is a norm-closed left ideal in $\mathcal{A}$. The sesquilinear form $\left<\cdot,\cdot\right>:\mathcal{A}/N_{\phi}\times\mathcal{A}/N_{\phi}\to\mathbb{C}$ defined by $\left<a+N_{\phi},b+N_{\phi}\right>:=\phi(b^*a)$ is a well-defined inner product on $\mathcal{A}/N_{\phi}$. The completion of $\mathcal{A}/N_{\phi}$ establishes a Hilbert space.

False belief: The completion is in the quotient norm.

Surprisingly, Wikipedia (as of April 27, 2018) presents a false statement "The quotient space of the A by the vector subspace I is an inner product space. The Cauchy completion of A/I in the quotient norm is a Hilbert space, which we label H."(https://en.wikipedia.org/wiki/Gelfand%E2%80%93Naimark%E2%80%93Segal_construction#The_GNS_construction) First of all, the quotient of a Banach space by its closed subspace is again a Banach space in the quotient norm, which is a very elementary fact in functional analysis. Thus A/I is already complete in the quotient norm, and hence there is no need to complete it in the quotient norm!

The correct completion is, of course, in the norm induced by the inner product, and this norm is not equivalent to the quotient norm in general. In fact, let $\mathcal{H}$ be a separable infinite-dimensional Hilbert space and $\{\xi_n\}_{n=1}^{\infty}$ be an orthonormal basis for $\mathcal{H}$. The linear functional $\phi:\mathbb{B}(\mathcal{H})\to\mathbb{C}$ defined by $\phi(a):=\sum_{n=1}^{\infty}\frac{1}{2^n}\left<a\xi_n,\xi_n\right>$ is a state on $\mathbb{B}(\mathcal{H})$, and $N_{\phi}=\{0\}$. Let $\xi_k\otimes\xi_k$ be the canonical rank-one operator, and put $p_n:=\sum_{k=1}^n\xi_k\otimes\xi_k$. Then $(p_n)_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{B}(\mathcal{H})/N_{\phi}$ in the norm induced by the inner product defined at the beginning, but it is NOT a Cauchy sequence in the quotient norm.

I have never seen a remark which clearly states the distinction between the norm induced by the inner product and the quotient norm in the literature on $C^*$-algebras. Since a quotient space is involved, students are easily tempted to think that the completion is in the quotient norm. (Even the Wikipedia editor was confused!) Or, they may thoughtlessly assume that these two norms are the same. So it will be instructive to clearly state the distinction between these two norms when one teaches this subject to undergraduate students.

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    $\begingroup$ I think it is surprising to be surprised by a wrong statement on Wikipedia. Fortunately it need not remain wrong! $\endgroup$
    – LSpice
    Apr 27, 2018 at 22:31
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    $\begingroup$ @LSpice: Well, the surprising thing is that this false statement appeared on May 6, 2004 and has remained since then, and nobody has corrected the error for 14 years! See the editing history. $\endgroup$ Apr 27, 2018 at 22:55
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    $\begingroup$ I read the "quotient norm" statement as saying that it's the quotient of the seminorm induced by the inner product. Then it's correct, no? $\endgroup$ Apr 28, 2018 at 12:24
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    $\begingroup$ @MasayoshiKaneda: I agree that the Wikipedia entry has been ambiguous at that point, and I've clarified it. $\endgroup$ Apr 29, 2018 at 7:14
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    $\begingroup$ @Tobias Fritz: Good job! I also modified the sentence before the one you clarified. $\endgroup$ May 1, 2018 at 10:51
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The derived subgroup of a finite group equals to the set of all its commutators

or equivalently

A product of two commutators in a finite group is always a commutator

This mistake is very widespread, probably because counterexamples to it tend to be quite large. The smallest group, for which it is not true has order $96$.

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    $\begingroup$ What is your evidence that this is a commonly held belief? $\endgroup$ Aug 28, 2020 at 12:06
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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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If we regard a ring $R$ (with identity) as a right module ($R_{R}$), then there is a ring isomorphism $\text{End}(R_{R}) \simeq R$, however the same does not happen if we regard $R$ as a left module!

The correct is $\text{End}(_{R}R) \simeq R^{\text{op}}$.

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  • $\begingroup$ Here is a discussion about the condition for Morita equivalence between rings, which is related to this subtle detail: math.stackexchange.com/questions/3566579/… $\endgroup$
    – user144185
    Mar 27, 2020 at 12:19
  • $\begingroup$ But this is just a notational quirk. In some Russian algebra texts for example, the composition "f then g" in $\operatorname{End}_R({}_RR)$ is written $fg$, which leads to $R$ being isomorphic, as a ring, to its (left $R$-linear) endormorphism ring. $\endgroup$
    – Jo Mo
    Jan 27 at 16:11
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Here's one that I think will surprise some number theorists:

False belief. Let $E$ be an elliptic curve over an algebraically closed field $k$ of characteristic $p > 0$. Then $\operatorname{End}^\circ(E)$ is strictly larger than $\mathbb Q$.

While this is true for all elliptic curves defined over finite fields, most elliptic curves whose field of definition is transcendental over $\mathbb F_p$ have $\operatorname{End}^\circ(E) \cong \mathbb Q$. The extra automorphism on elliptic curves over a finite field comes from the geometric Frobenius. For varieties over larger fields, this is not a thing.

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The "curse of dimensionality" means that in a hypercube the volume is increasingly concentrated in the corners as the number of dimensions increase.


In fact half the volume of a hypercube is closer to the centre than to the nearest vertex, with any number of dimensions.

The real curse is that the vast majority of the points of a unit hypercube of dimension $n$ are a distance less than $\frac{5}{n}$ from the outside of the hypercube, distances $\sqrt{\frac{n}{12}} \pm \frac12$ both from the centre and from the nearest vertex, and a distance $\sqrt{\frac{n}{6}} \pm 1$ from the vast majority of other points, which for large $n$ are narrow bands.

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Given a finite dimensional vector space $V$ over $\mathbb{R}$ it cannot be written as a countable union of proper subspaces. (This can be proved by algebraic arguments or by the Baire category theorem.)

This may lead one to believe that the same is true if $V$ is infinite dimensional. However, that is false!

The vector space $P$ of polynomials with real coefficients is the union of the subspaces $P_n$ of polynomials of degree $n$.

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    $\begingroup$ I like the example, but "This may lead one to believe" is not the same as "this is a common false belief". $\endgroup$ Jan 26 at 11:16
  • $\begingroup$ True enough! However, I have often seen people think that throwing in some additional conditions will make it true. For example, I have heard the assertion, "In a normed space a countable union of closed subspaces cannot be the whole space." This is also false with the same counter-example and a norm like the sup norm. $\endgroup$
    – Kapil
    Jan 26 at 11:29
  • $\begingroup$ Of course, throwing in enough additional conditions does make it true. No infinite-dimensional Bach space is the countable union of closed proper subspaces. $\endgroup$ Apr 13 at 20:41
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From Kleiman's article "Misconceptions about $\mathcal{K}_X$", the sheaf of meromorphic functions:

Denote by $A_\mathrm{tot}$ the total fraction ring of a ring $A$.

(1) $\mathcal{K}_X$ can be defined as the sheaf associated to the presheaf of total fraction rings $$U \mapsto \Gamma(U, \mathcal{O}_X)_\mathrm{tot}$$

(2) The stalks of the meromorphic functions are the total fraction rings of the stalks: $$\mathcal{K}_{X,x} = (\mathcal{O}_{X,x})_\mathrm{tot}$$

(3) If $U = \mathrm{Spec}(A) \subset X$ is an affine open, then $$\Gamma(U,\mathcal{K}_X) = A_\mathrm{tot}$$

The first two misconceptions apply to any ringed space $X$, and the third applies to a scheme. Please see his nice, three-page article for discussion and examples.

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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!) $\endgroup$ Oct 4, 2010 at 19:21
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Let $R$ be a ring with identity $e$, $A, B\in R$, $A\neq 0$, $B$ is invertible element. If $A\cdot B = A$ then $B = e$.

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  • $\begingroup$ I think, it is closely related to the following false "deduction": because invertible element cannot be at the same time zero divisor, therefore sum of any unit and zero divisor is not invertible. Ok, maybe it isn't popular, but I've got this belief at my first algebra course, until I discovered counterexample $1+X$ in $R[X]/X^2$. This is almost exactly the thing you mentioned, just put $B:=X, A:=X+1$. $\endgroup$
    – Przemek
    Nov 5, 2020 at 20:09
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Assume that $a,b\in \mathbb{R}\setminus \{0\}$ which satisfy $a^{3}= 2b^{3}$.

Then $a-2b$ is a non zero nilpotent element of group ring $\mathbb{Z}_{3} \mathbb{R}$, that is $(a-2b)^{3}=0$.

This would be a counterexample to the zero divisor Kaplansky conjecture

The false lies in an obvious abuse in the definition of the group ring multiplication.

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    $\begingroup$ This does not seem like a common false belief. $\endgroup$
    – Yemon Choi
    May 15, 2016 at 11:47
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Let $M \subset B(H)$ be a von Neumann algebra, $p \in B(H)$ a projection and $q=I-p$.

False belief: If $pM=Mp$ then $M=pMp \oplus qMq$.
(I think it is a quite common careless mistake)

Counter-example: diagonal embedding of $\mathbb{C}$ into $M_2(\mathbb{C})$.

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I don’t know how common is the following false belief, but I had it for several years, so maybe some other people also have it. I apologize to those to whom I shared this false belief. I hope this post will help.

Kaplansky’s 6th conjecture (here, 1975) states that if $H$ is a finite dimensional semisimple Hopf algebra and $V$ an irreducible representation of $H$, then $\dim (V)$ divides $\dim (H)$. This conjecture is open over the complex field $\mathbb{C}$, but false in positive characteristic. So we assume to be over $\mathbb{C}$.

For the group case, this property was proved by Frobenius, that is why a finite dimensional semisimple Hopf algebra (over $\mathbb{C}$) with this property is called of Frobenius type.

A finite dimensional Hopf algebra (over $\mathbb{C}$) is called a finite quantum group (or Kac algebra) if it has a $*$-structure. And then it is also semisimple. It is an open problem whether such a $*$-structure always exists.

False belief: George Kac proved Kaplansky’s 6th conjecture for the finite quantum groups.

This false belief was pointed out to me by Pavel Etingof after this talk I gave for Harvard University, and where I mentioned it. Fortunately, that does not affect the content of the talk.

What I had in mind is Theorem 2 in the following paper:
G. I. Kac, Certain arithmetic properties of ring groups., Funct. Anal. Appl., 6 (1972), pp. 158–160.

In modern language, Theorem 2 proves the following: let $H$ be a finite quantum group, and let $\mathcal{C} = Corep(H)$ be the fusion category of complex corepresentations of $H$. For every simple object $X$ of the Drinfeld center $Z(\mathcal{C})$ which contains the trivial object of $\mathcal{C}$ under the forgetful functor, $FPdim(X)$ divides $FPdim(\mathcal{C}) = \dim(H)$ (the quotients are called the formal codegrees).

Note that these $X$ correspond to the irreducible representations of the Grothendieck ring $K(\mathcal{C})$ of $\mathcal{C}$ (see Theorem 2.13 here). In particular, for $G$ a finite group, $\mathcal{C} = Corep(G) = Vec(G)$, and $Irr(K(\mathcal{C})) = Irr(G)$. That is why Theorem 2 implies Kaplansky’s 6th conjecture in the group case (i.e. covers Frobenius theorem). But it is not clear for a finite quantum group in general. It could be relevant to search in this direction, in particular to check whether for every object $Y$ of $Irr(H)$ there exists an $X$ as above such that $\dim(Y)$ divides $FPdim(X)$, because this would prove that $H$ is a Frobenius type.

Note that Theorem 2 (as stated above) holds more generally for every (complex) fusion category $\mathcal{C}$. The case $\mathcal{C} = Rep(G)$, with $G$ a finite group, recovers the fact that the size of each conjugacy class of $G$ divides $|G|$. Finally, according to Pavel, the theorem holds more generally without the assumption ‘which contains the trivial object’ (I don’t have the exact reference for that, so if you know it, please put it in comment).

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A common false assumption is that that two non-orthogonal pure states of a quantum mechanical system may never be unambiguously distinguished by a measurement. (See https://arxiv.org/pdf/quant-ph/9807022.pdf)

Another false belief is that a quantum computer is similar to an analogue computer, in that large computations will necessarily fail because of accumulated error. (See, for example, https://arxiv.org/abs/quant-ph/9712048)

For that matter, another common false believe is that Bell Inequalities aren't violated, although it is mostly held by people who have never heard of Bell Inequalities.

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    $\begingroup$ I'm not sure how you can believe that something you have never heard of isn't violated. $\endgroup$ Apr 10, 2016 at 17:55
  • $\begingroup$ @GeoffRobinson: I have on remarkably many occasions introduced good mathematicians to the Bell inequalities and been told that it's clearly impossible for them to be violated. $\endgroup$ Jan 5 at 22:12
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    $\begingroup$ @StevenLandsburg: But by then they have heard of them! $\endgroup$ Jan 5 at 23:02
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If $R$ is a commutative ring with $char(R)=p$ ($p$ is prime), then

$$\varphi:R \to R$$ $$x\mapsto x^p$$

is an automorphism.

Which is false of course.

(If $R$ is a field then see comments).

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  • $\begingroup$ @SamHopkins Pay attention, I changed the statement because accidentally I wrote field instead of ring $\endgroup$
    – Or Shahar
    Apr 13 at 19:45
  • $\begingroup$ But even with field it is still false? $\endgroup$ Apr 13 at 19:47
  • $\begingroup$ @SamHopkins Well, if $R=\Bbb{F}_p$ then $x^p$ is an automorphism because it’s a 1 to 1 on a finite set and endomorphism. If the field is infinite, then take $\Bbb{F}_p(x)$, there isn’t exist $f(x)\in \Bbb{F}_p(x)$ such that $f(x)^p=x$ $\endgroup$
    – Or Shahar
    Apr 13 at 20:02
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    $\begingroup$ Yes. I guess I just meant that since the (false) assertion is "weaker" with field instead of ring, as a false belief it might be more common. $\endgroup$ Apr 13 at 20:03
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I don't know how common this is, but it occurs as a corollary of a theorem in the fine, and widely used, text by Shafarevich on algebraic geometry: namely, if $f \colon X \longrightarrow Y$ is a surjective algebraic map of varieties, then 1) for all $y \in Y$, the fiber over $y$ has dimension $≥ \dim(X)-\dim(Y)$; 2) on some non empty open set in $Y$ the dimension of the fibers equals $\dim(X)-\dim(Y)$; 3) for all $r$, the set of $y \in Y$ such that the fiber over $y$ has dimension $≥ r$, is closed in $Y$.

The first two are true, but the third is false. Upper semicontinuity of fiber dimension is true on the source, not the target. For the conclusion as stated to hold, one can add properness to the hypothesis on the map. I think this is not at all widely believed by experts, but for some reason it persists in the text, hence may be believed by students.

Since I have myself written notes in which blatantly false statements occur, I do not think for a moment that Shafarevich himself believed this false statement. But such things do slip by, and may mislead beginners. In fact I believed it for some time until enlightened by a friend.

In keeping with the OP's desire to know the psychological reason for the error, it seems for some reason common in my experience for people to assume unconsciously that maps are proper.

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False belief: a subgroup isomorphic to a quotient is a retract.

Formally: Let $H,N$ be subgroups of $G$ with $N$ normal and $H \simeq G/N$, then $H$ is a retract of $G$.

It is false, because otherwise $C_2$ would be a retract of $C_4$, but it is not.

In fact, $H$ is a retract of $G$ if and only if $G$ is isomorphic to $H \ltimes N$ (semidirect product).

This false belief caused this post.

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I cannot believe this example was not yet given (but, if my belief is false, I will happily delete this answer):

It is very common among "lay people" (who do not understand what it means for lines to be parallel) to believe that "in some kind of geometry" (frequently described as non-euclidean) parallel lines can intersect.

One finds many instances (my guess, the count is in hundreds of thousands) of this false believe just by searching the internet. Here is a random example, the article "How Looking At A Basketball Disproves Something Everybody Learns In High School Geometry" from "Business Insider", 2014. The article concludes with

And voila! We’ve successfully disproven “parallel lines never intersect” using just a basketball.

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    $\begingroup$ i.redd.it/r7etb5kayl961.jpg $\endgroup$ Feb 9, 2021 at 4:39
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    $\begingroup$ I feel like this is a confusion of terminology and not mathematics. The definition of parallel being used in the Business Insider article is clearly not 'lines that do not meet and are a constant distance apart' or whatever the official definition of parallel is. It is closer to 'distinct lines that go in the same direction'. Insofar as there is a false belief, it is that there is a coherent notion of 'same direction'! But this is not essential to the belief that lines that go in the 'same direction' can meet in some kind of geometry, which is a belief that is closer to true than false. $\endgroup$
    – Chan Bae
    Feb 9, 2021 at 13:37
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    $\begingroup$ @ChanBae Mathematics starts by establishing common terminology and axioms. This is what Greeks realized over 2000 years ago. Sadly, this understanding was lost with changes in math education in the last century. The thing is, math is part language and part science. You cannot separate the two and claim that inability to understand definitions is just a matter of terminological disagreements. $\endgroup$ Feb 10, 2021 at 1:37
  • $\begingroup$ @MoisheKohan : It is not unusual in mathematics for the same word to have several definitions depending on context, and I think all that is happening here is that people are using the word "parallel" in a perfectly well defined way that they fully understand and that happens to be different than the way you had in mind. Some people (including me) believe that a tangent to a circle is normal to the diameter. I do not think that counts as a false belief in mathematics just because "normal" also means "fixed under conjugations". $\endgroup$ Jan 6 at 1:48
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If an Abelian category $\mathcal{A}$ is a full subcategory of an Abelian category $\mathcal{B}$, then for all objects $M,N$ of $\mathcal{A}$, we have an injection $$\operatorname{Ext}^i_{\mathcal{A}}(M,N) \hookrightarrow \operatorname{Ext}^i_{\mathcal{B}}(M,N).$$

As an example, let $G$ be the free group on $2$ letters, $A$ its abelianization, $\mathcal{B} = G-mod$, $\mathcal{A}=A-mod$, and $M=N=\mathbb{Z}$ with the trivial action. Then $\operatorname{Ext}^i_{\mathcal{A}}(M,N) \cong \mathbb{Z}$, while $\operatorname{Ext}^i_{\mathcal{B}}(M,N) \cong 0$.

(This example comes from the topological fact that a torus has nontrivial $H^2$, while a punctured torus has trivial $H^2$. In algebra, it's related to the idea that group homology $H_1$ is space of generators for a group while $H_2$ is a space of relations.)

This false belief came up a context where $\mathcal{B}$ was the category of all Galois representations while $\mathcal{A}$ was a certain subcategory. See the comments to Status of the conjectured vanishing of Bloch-Kato H^2.

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Another common mistake. If $W = _P(e_1,\ldots, e_{n})$ is a vector space and $V$ is a subspace of $W$ of dimension $k$, then $V = _P(e_{i_1},\ldots, e_{i_k})$.

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  • $\begingroup$ What does that little subscript $p$ on the equals sign mean? $\endgroup$ Feb 11, 2016 at 21:36
  • $\begingroup$ $V$ is a vector space over field $P$. $\endgroup$ Feb 12, 2016 at 6:52
  • $\begingroup$ So, what does "$V$ is a vector space over field $P$ $(e_1,\dots,e_n)$" mean? $\endgroup$ Feb 12, 2016 at 8:37
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    $\begingroup$ $W$ is a vector space over field $P$, $(e_1,\ldots, e_n)$ is a basis of $W$. $V$ is a subspace of $W$. $\endgroup$ Feb 12, 2016 at 8:40
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Initially when I started studying sequences I believed that:

Consider $(x_n)$ and $(y_n)$ are two convergent real sequences having limits $x$ and $y$ as limits respectively. If $ x_n < y_n \quad \forall n\in \mathbb{N}$ then $x < y$

which turned out to be false.

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    $\begingroup$ but the question is "less interested in very elementary false statement like $(x+y)^2=x^2+y^2$ " $\endgroup$ Aug 16, 2021 at 9:21
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