Examples of common false beliefs in mathematics

Question

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.

Answer 1

(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 ?" .

Answer 2

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.

Answer 3

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]]).

Answer 4

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 $P^A=NP^A$.

Answer 5

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.

Answer 6

I already thought that the following two sets of matrices are one. $$M(\color{blue}{\Bbb R},2n)\qquad \text{and}\qquad M(\color{red}{\Bbb C},n).$$

Answer 7

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 $n$.

Answer 8

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.

Answer 9

Cambridge mathematicians (with the notable exception of the exceptional Dirac) have often misunderstood the Euler-Heaviside fractional integro-derivative calculus, disadvantaging their students.

False belief: A erroneous claim that has been often repeated is that the fractional calculus (FC) as envisioned by Euler and Heaviside (Hadamard, Pincherle, and others) doesn't obey the law of exponents

$D^{\alpha}D^{\beta} =D^{\alpha+\beta},$

and, specifically, differentiation $D$ and integration $D^{-1}$ do not commute, and consequently, neither do they obey the law of exponents.

One example of this apparent lack of commutativity is given on the webpage Fractional Calculus III (circa 2008) by Beardon:

$$D^{-1}D \; e^x = \int_0^x e^t dt = e^x-1 \neq DD^{-1} e^x = D(e^x-1) = e^x = D^0e^x = D^{-1+1}e^x.$$

In fact, however, commutativity applies once the Heaviside step function $H(x)$ is correctly introduced since

$$D^{-1}D \; H(x) e^x =H(x) \int_0^x (\delta(t) +e^t) dt =H(x)( 1+ e^x-1) =H(x) e^x = D^0H(x) e^x .$$

The mistaken claim of lack of commutativity is repeated in another example by Beardon illustrated below (and similar examples by others).

More generally, the Euler-Heaviside FC can be framed a number of ways to ensure the fundamental operator action is interpreted as

$$D^{\alpha}D^{\beta} H(x) \frac{x^{\gamma}}{\gamma!} =D^{\beta} D^{\alpha} H(x) \frac{x^{\gamma}}{\gamma!} =D^{\alpha+\beta}H(x) \frac{x^{\gamma}}{\gamma!} = H(x) \frac{x^{\gamma-\alpha-\beta}}{(\gamma-\alpha-\beta)!}$$

for $\alpha,\beta,$ and $\gamma$ any real numbers.

In one interpretation (e.g., see Gelfand and Shilov's Generalized Functions Vol. I, p. 57),

$$H(x) \frac{x^{-n-1}}{(-n-1)!}= D^n \delta(x)= \delta^{(n)}(x)$$

such that, under a finite part construction or other analytic continuation,

$$D^{n} H(x)\frac{x^{\alpha}}{\alpha!} =H(x) \int_0^x \frac{(x-t)^{-n-1}}{(-n-1)!} \frac{t^{\alpha}}{\alpha!}dt = H(x)\oint_{|z-x|=x} \frac{n!}{(z-x)^{n+1}}\frac{z^{\alpha}}{\alpha!}dz =H(x) \frac{x^{\alpha-n}}{(\alpha-n)!}$$

and, more generally,

$$D^{\beta} H(x)\frac{x^{\alpha}}{\alpha!} = H(x)\int_{-\infty}^\infty H(x-t) \frac{(x-t)^{-\beta-1}}{(-\beta-1)!} H(t)\frac{t^{\alpha}}{\alpha!}dt$$

$$= H(x)\int_0^x \frac{(x-t)^{-\beta-1}}{(-\beta-1)!} \frac{t^{\alpha}}{\alpha!}dt = H(x)\oint_{|z-x|=x} \frac{\beta!}{(z-x)^{\beta+1}}\frac{z^{\alpha}}{\alpha!}dz =H(x) \frac{x^{\alpha-\beta}}{(\alpha-\beta)!}.$$

Then the Euler-Heaviside FC gives

$$D^{\frac{1}{2}} H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!} = H(x)\frac{x^{-1}}{(-1)!} = \delta(x)$$

and

$$D^{\frac{1}{2}}D^{\frac{1}{2}} H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!}$$$$ = D^{\frac{1}{2}} H(x)\frac{x^{-1}}{(-1)!} = H(x)\frac{x^{\frac{-3}{2}}}{(\frac{-3}{2})!} = D H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!},$$

so, in this case,

$$D^{\frac{1}{2}}D^{\frac{1}{2}} = D$$

whereas Beardon concludes that

$$D^{\frac{1}{2}}x^{\frac{-1}{2}} = 0,$$

implying that the law of exponents is violated since then

$$D^{\frac{1}{2}} D^{\frac{1}{2}}x^{\frac{-1}{2}} = D^{\frac{1}{2}}0 = 0 \neq D x^{\frac{-1}{2}} =\frac{-1}{2}x^{\frac{-3}{2}} .$$

H. and B. Jeffreys on p. 229 of their book Methods of Mathematical Physics (Second Ed., 1950), often referenced in discussions of operational calculus, assert unqualifiedly that integration and differentiation do not commute. And, Lavoie, Osler, and Trembley in "Fractional derivatives and special functions" (1976) repeated the false argument of the unqualified violation of the law of exponents. A lack of appreciation of the roles of the Heaviside step and delta functions in the Euler-Heaviside FC seems to be at the core of this oversight with a concomitant tendency (issuing from some blend of laziness, carelessness, and quasi-authoritarianism) to reprint the claims of previous researchers who have omitted discussions of the core constructs--the step and delta functions--in their analyses.

If the FC is constructed using an infinitesimal generator, the analytic continuations can be dodged, or a Pochhammer contour integral can be invoked for generalizing the beta function integral. These constructions are consistent with Laplace- and Mellin-transform approaches over the domains of common convergence and analytic continuation of the reps, with Pincherle's axiomatic treatment of a canonical FC, and with the calculus of Appell Sheffer polynomial sequences.

Mikusinski provided an analogous algebraic convolutional approach, and Sato, a hyperfunction approach, reflecting earlier presentations by Niels Nielsen and then much later than Nielsen by Feynman and others.

(An equally erroneous claim is made to the opposite extreme in an answer to this MO-Q that differentiation and integration operators commute unqualifiedly.)

Answer 10

Just today, I realised that I had been mis-interpreting the FTFGAG. One can speak of the $2^\infty$-torsion subgroup or the $2'$-torsion subgroup, the $3^\infty$-torsion or the $3'$-torsion subgroup, or even just the torsion subgroup of a FGAG or the maximal torsion-free quotient … so surely one can speak of the torsion-free subgroup, right? In fact, when I was corrected on this, my first thought was to reply: "just take the subgroup consisting of all infinite-order elements", and it only occurred to me as I was saying it to wonder how the identity element would squeeze its way into this so-called 'subgroup'.

Answer 11

I’m not sure if this counts as “reasonably advanced”, but given the wide-ranging nature of the answers here, I feel like this is worth a mention, given that it plagued my mind for so long when I was in high school.

$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ for all real numbers $a, b$.

Of course, what exactly does $\sqrt{a}$ even mean when $a < 0$ is not completely clear. I’ll just take $\sqrt{a} = i\sqrt{|a|}$ for negative $a$, which seems perfectly reasonable given that we usually teach students $i = \sqrt{-1}$. Under this definition, the above statement is false. Indeed, $\frac{\sqrt{1}}{\sqrt{-1}} = \frac{1}{i} = -i$ but $\sqrt{\frac{1}{-1}} = \sqrt{-1} = i$. When I first encountered complex numbers, I obtained a “proof” that $1 = -1$ by multiplying both sides of $i = \sqrt{-1} = \sqrt{\frac{1}{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \frac{1}{i}$ by $i$ and I had been wondering all throughout my high school years where the error was in the proof until I worked out why exactly I thought $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ and figured out it doesn’t work for negative $a$ or $b$.

Answer 12

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

https://en.wikipedia.org/wiki/Matrix_exponential#Evaluation_by_Laurent_series (current revision)

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 $\mathbb C[[X-s]]$.

---

Answer 13

A direct sum of simple algebras is semisimple.

This is the standard reformulation for the finite-dimensional case. If we define a semisimple algebra as an associative Artinian algebra over a field with a trivial Jacobson radical, as stated in the Wikipedia link, this property does not generally apply. Specifically, the Artinian condition is not met when the direct sum involves infinitely many simple algebras.

In the definition we are using, semisimple algebras must be Artinian. However, some authors use semisimple more loosely to refer to algebras that simply have a trivial Jacobson radical, without needing to be Artinian. On Wikipedia, this is called semiprimitive or Jacobson semisimple.

Answer 14

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" :)

Answer 15

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.

Answer 16

The assumption that a cubic surface expressed as a foliation of Weierstrass curves cannot be rational, because a general Weierstrass curve is not rational.

I've seen this false assumption more than once on sci.math over the years. But there are simple counterexamples, such as:

$ (x + y) (x^2 + y^2) = z^2 $

On defining $ u = x/y $ and $ v = z/y $ one obtains $ y (u + 1) (u^2 + 1) = v^2 $, and hence x, y, z as rational functions of u, v.

I'd love to have a reference to a procedure for calculating the geometric genus and algebraic genus of surfaces like this, because they are rational if and only if both these quantities are zero, and for other cubic surfaces that interest me it would save a lot of fruitless hacking around trying to find a rational solution that probably doesn't exist! Are there any symbolic algebra packages that can do this?

I mean for example is $ x y (x y + 1) (x + y) = z^2 $ rational? I'm almost sure it isn't; but how can one be sure?

Answer 17

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 $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$.

Answer 18

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

Answer 19

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 $W_\kappa$.

Answer 20

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.

Answer 21

Anytime I wanted to write an answer to this question, I doubted maybe it is not as common as worthy of mentioning here. In fact, I am also not sure how common is the false belief that I observed today in a PDE class. I didn't observe that in many years of teaching calculus, but today four or five students in a small PDE class when calculating a definite integral by parts only applied the limits of the integral to the "second" integral, that is:

$$\int_{a}^b{f(x) g'(x) dx}=f(x) g(x) - \int_{a}^b{f'(x) g(x) dx}$$

Haven't I observed well enough in my calculus classes?

Answer 22

This might not be common, but I once believed the following.

Let $ A, B $ be integers, and define a sequence by the linear recurrence $ s_n = A s_{n-1} + B s_{n-2} $ with the base case $ s_0 = 0 $, $ s_1 = 1 $. Two important special cases are the Fibonacci sequence ($ A = B = 1 $) and the sequence $ s_n = 2^n - 1 $ (where $ A = 3 $, $ B = -2 $). Then, for any integers $ n $ and $ k $, $ \gcd(s_n, s_k) = s_{\gcd(n,k)} $.

This is true in the two mentioned special cases, so it's tempting to believe it's true in general. But there's a counterexample: $ A = B = k = 2 $, $ n = 3 $.

Update: corrected the powers of two minus one example from $B = 2$ to $B = -2$. Thanks to Harry Altman.

Answer 23

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 $96$.

Answer 24

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 $T_2\subseteq T_1$.

Answer 25

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).

Answer 26

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.

Answer 27

“A reversible computer can factor integers efficiently in polynomial-time”: Since a reversible computer can multiply two integers efficiently in polynomial-time, it can also factor an integer efficiently in polynomial-time by just putting the computer in reverse.

I wish this were true.

Answer 28

A dense subspace of a Hilbert space $H$ must contain an ONB for $H$.

This is, of course, true if $H$ is separable, but false in general. See, for example, this answer in MSE: https://math.stackexchange.com/a/201149/465145.

Answer 29

Root(s) of a $3^\text{rd}$ degree polynomial over $ \Bbb Q$ are expressible using radicals with the imaginary $i$. If a root $r$ is real, by taking only the real part, $r$ is expressible using radicals over the rational numbers.

Why not? See Casus irreducibilis on Wikipedia.

Answer 30

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.