Q: In your opinion, which fundamental theorem is incredibly missing??…

Fundamental Theorem of Arithmetics

Fundamental Theorem of Algebra

First Fundamental Theorem of Calculus

A: Here we just have uniqueness for the answer!

The Fundamental Theorem of Arithmetics

The Fundamental Theorem of Arithmetics – Why isn’t the fundamental theorem of arithmetic obvious?The fundamental theorem of arithmetic states that every positive integer can be factorized in one way as a product of prime numbers. This statement has to be appropriately interpreted: we count the factorizations and as the same, for instance. Note that it is essential not to count 1 as a prime, or else we could stick a product of 1s on to the end of any factorization to get a different one: . But doesn’t that mean that 1 itself cannot be written as a product of primes? No — we define the “empty product” (what you get when you take a bunch of … no numbers at all and multiply them together) to be 1. That is a sensible convention because we would like multiplying a product of numbers by the empty product not to make any change to the result.

That’s enough about what the fundamental theorem of arithmetic says. In this post I want to discuss the question of why it is a theorem at all. Isn’t it more like an

observation? After all, given any number, we can simply work out its prime factorization.

Answer 1.If you think it’s obvious, then you’re probably assuming what you need to prove.If you say, “we can simply work out its prime factorization,” you are already assuming that that factorization is unique. Otherwise, you would have had to say, “we can simply work out a prime factorization for it”. Of course, if you say it that way, it suddenly doesn’t seem quite as obvious that there’s only one. If you’re trying to argue that it’s obvious and you ever utter the phrase, “the prime factorization,” then you are begging the question, since implicit in those words is the assertion that there is only one prime factorization.

Answer 2.Just because you’ve got a completely deterministic method for working out a prime factorization, that doesn’t mean what you work out is the only prime factorization.The following method is probably how you factorize a number: you divide it by 2 as many times as you can (which may be no times at all), then by 3, then by 5, and so on, keeping track of what you’ve done. For example, if your starting number is 575, then you can’t divide it by 2 or 3, but you can divide it by 5 to get 115, and then by 5 again to get 23, and then … well, you’ll probably know that 23 is prime, but you could also argue that since is greater than 23 and you’ve checked 2 and 3, then it

mustbe prime.But just because that

methodalways gives the same answer (the method in the abstract being to keep dividing by the smallest prime that goes into the number you see in front of you), that doesn’t mean that there might not be someothermethod that gives a different answer. For example, what if you looked for thelargestprime that went into your number? You’re probably thinking that you’ll just get the same list of primes, but written backwards. But how do youknowthis? Obviously that’s what you’ll get if there’s only one way of writing the number as a product of primes, but that’s what we’re trying to prove. If there’s another way of writing it as a product of primes, then perhaps the largest prime in the other way of doing things is larger than the largest prime that results from the usual method.

Answer 3.Look, it just bloody well isn’t obvious, OK?Sorry, I lost it for a moment there. But if you persist in thinking that it’s obvious, then perhaps you can tell me why it is obvious that is not the same number as . I’ll save you a little time by revealing that all of 23, 53, 769 and 1759 are prime. I will not accept as an answer that if you calculate those two products you get different results. That to me is an admission that it

wasn’tobvious that the answers would be different. If it was obvious, then why bother to calculate them?By the way, I’ll grant you that sometimes it’s obvious that two products of primes are different. For example, if 2 is involved in one product and not the other, then the first product is even and the second is odd. However, even that second assertion depends on the (simple) result that a product of odd numbers is odd. We’d be able to see instantly that if we knew that a product of two non-multiples of 23 was always a non-multiple of 23. But is there an easy way of showing that? We can work out the multiplication table mod 23, but that’s a bit tedious. Alternatively, we can use some theory from the course — but unless you’re finding the course so easy that that theory (a proof derived from Euclid’s algorithm) is utterly obvious, then I don’t think you can call it obvious that .

Here’s another pair of products of primes for your delectation and delight: and . Are they obviously different? It’s not clear which is bigger — they’re both a little over 40,000. What about the last digit? Damn, 1 in both cases. OK, let’s go for the second last digit, which is a bit of a cheat but still. In the first case we get the last digit of , which is 6. In the second case we get the last digit of , which is again 6. So we’ve got two numbers that are a little bit above 40,000 that both end 61. As it happens and .

If you wanted a quicker demonstration that those two numbers are different, you could work out what they are mod 3, which is a lot easier than working them out completely. But that’s not going to work in general. For instance, it doesn’t work for the first example, where the smallest modulus for which they differ is 7. If I took two pairs of absolutely huge numbers (with millions of digits), I could get them to agree in almost all their digits and differ by a multiple of, say, 1000! And even if such small-modulus tests can be used, it isn’t obvious in advance that they will work if it isn’t obvious in advance that the products are different.

Answer 4.If it’s so obvious that every number has a unique factorization, then why is the corresponding statement false in a similar context?Consider the collection of all numbers of the form where and are integers. (You might prefer to write these numbers as , but I prefer for reasons that I don’t want to go into here, but might mention in a future post.)

These numbers have various properties in common with the integers: you can add them and multiply them, there are identities for both addition and multiplication, and every number has an additive inverse. And as with integers, if you divide one by another, you don’t always get a third, so the notion of divisibility makes sense too. That means that we could if we wanted try to define a notion of a “prime” number of the form .

Just before I try to do that, let’s quickly decide what we mean by a prime when we allow negative numbers. Presumably we’re going to want, say, to be a prime, but what definition will lead to that? The small technical obstacle we face is that if we allow negative primes like that, then for a somewhat silly reason factorizations won’t be unique: for instance, . The usual approach to this is to divide numbers into three kinds: prime numbers, composite numbers, and

units. A unit is a number that has a multiplicative inverse, so in the units are 1 and -1. A prime is a number that cannot be written in the form unless exactly one of and is a unit. (I said “exactly” one because I didn’t want accidentally to define units themselves to be primes.) And now we can express the fundamental theorem of arithmetic in by saying that every number has exactly one factorization into primes, except that we count two factorizations as the same if the only difference (apart from the order) is that the primes in one factorization are multiplied by units to give the primes in the other factorization. For example, we count and as the same, since we can reorder the second factorization as and then multiply both primes by the unit -1 to get , which gives us the first factorization.In short, what we’re saying is that if two products of primes don’t obviously give the same number, then they give different numbers.

Right, back to numbers of the form . Let’s check that 2 is a prime in this ring. (A

ringin this context is, roughly speaking, an algebraic structure with addition and multiplication with all the usual axioms apart from the existence of multiplicative inverses. You can think of it as something a bit like . However, the actual definition is a bit more general, as you can find out from the relevant Wikipedia article. Hmm, I’ve just looked at that article and I don’t like it at all: the list of examples is woefully inadequate. The important examples are eventually mentioned, but not in the list of basic examples, so you don’t get a good idea that they are the important ones.) First of all, 2 isn’t a unit, since is not of the form . The modulus of is , so if , then the modulus of is bigger than 2. It follows that the only way of writing 2 as a product of non-units would have to be to write it as a product of non-unit integers, which we can’t. So 2 is prime.A similar check can be run for 3. So 2 and 3 are primes. It’s also possible to show that and are primes. But , so 6 has a non-unique factorization into primes. (It’s also easy to see that you can’t multiply or by a unit to get one of .)

Why is this a problem for people who hold that the fundamental theorem of arithmetic is obvious? It’s because they have to explain what it is about that is relevantly different from the ring of numbers of the form , which is denoted . Why can’t we just translate any proof that works for into a proof that works for ?

Here’s an example of how you can use to defeat somebody who claims that the result is obvious in . Let’s take the argument that you can just work the factorization out by repeatedly dividing by the smallest prime that goes into your number. Well, you can do that in as well. Take 6, for instance. The smallest prime (in the sense of having smallest modulus) that goes into 6 is 2. Dividing by 2 we get 3, which is prime. So we’re done. So there can’t be another factorization. Except that there

isanother factorization. So the argument just isn’t an argument.In a future post I’ll discuss the proof of the fundamental theorem of arithmetic. But this post is just to try to convince you (if you needed convincing, which you may not have) that the result is worth going to some effort to prove.

## Source: Gower´s blog

Fundamental Theorem of Arithmetics

Euclid‘s Elements – Essentially the statement and proof of the fundamental theorem

Proposition 30 is referred to as Euclid’s lemma. And it is the key in the proof of the fundamental theorem of arithmetic:

If two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers.

— Euclid, Elements Book VII, Proposition 30

Proposition 31 is proved directly by infinite descent:

Any composite number is measured by some prime number.

— Euclid, Elements Book VII, Proposition 31

Proposition 32 is derived from proposition 31, and prove that the decomposition is possible:

Any number either is prime or is measured by some prime number.

— Euclid, Elements Book VII, Proposition 32

Book IX, proposition 14 is derived from Book VII, proposition 30, and prove partially that the decomposition is unique – a point critically noted by André Weil.^{[7]}Indeed, in this proposition the exponents are all equal to one, so nothing is said for the general case. Article 16 of Gauss‘ Disquisitiones Arithmeticae is an early modern statement and proof employing modular arithmetic:^{[1]}If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it.

— Euclid, Elements Book IX, Proposition 14

Existence ProofWe need to show that every integer greater than 1 is either prime or a product of primes. For the base case, note that 2 is prime. By induction: assume true for all numbers between 1 and

n. Ifnis prime, there is nothing more to prove. Otherwise, there are integersaandb, wheren=aband 1 <a≤b<n. By the induction hypothesis,a=p_{1}p_{2}…p_{j}andb=q_{1}q_{2}…q_{k}are products of primes. But thenn=ab=p_{1}p_{2}…p_{j}q_{1}q_{2}…q_{k}is a product of primes.

Uniqueness ProofAssume that

s> 1 is the product of prime numbers in two different ways:

We must show

m=nand that theq_{j}are a rearrangement of thep_{i}.By Euclid’s lemma,

p_{1}must divide one of theq_{j}; relabeling theq_{j}if necessary, say thatp_{1}dividesq_{1}. Butq_{1}is prime, so its only divisors are itself and 1. Therefore,p_{1}=q_{1}, so that

Reasoning the same way,

p_{2}must equal one of the remainingq_{j}. Relabeling again if necessary, sayp_{2}=q_{2}. Then

This can be done for each of the

mp_{i}‘s, showing thatm≤nand everyp_{i}is aq_{j}. Applying the same argument with the p‘s and q‘s reversed showsn≤m(hencem=n) and everyq_{j}is ap_{i}.

Canonical Representation of a Positive IntegerEvery positive integer

n> 1 can be representedin exactly one wayas a product of prime powers:

where

p_{1}<p_{2}< … <p_{k}are primes and the α_{i}are positive integers. This representation is commonly extended to all positive integers, including one, by the convention that the empty product is equal to 1 (the empty product corresponds tok= 0).

Arithmetic OperationsThe canonical representation, when it is known, is convenient for easily computing products, gcd, and lcm:

However, as Integer factorization of large integers is much harder than computing their product, gcd or lcm, these formulas have, in practice, a limited usage.

GeneralizationsThe first generalization of the theorem is found in Gauss’s second monograph (1832) on biquadratic reciprocity. This paper introduced what is now called the ring of Gaussian integers, the set of all complex numbers

a+biwhereaandbare integers. It is now denoted by Z [ i ] . He showed that this ring has the four units ±1 and ±i, that the non-zero, non-unit numbers fall into two classes, primes and composites, and that (except for order), the composites have unique factorization as a product of primes.^{[11]}In 1844 while working on cubic reciprocity, Eisenstein introduced the ring Z [ ω ], where

satisfy , i.e. is a cube root of unity. This is the ring of Eisenstein integers, and he proved it has the six units

and that it has unique factorization. However, it was also discovered that unique factorization does not always hold. An example is given by Z [ − 5 ]. In this ring one has

^{[12]}

Examples like this caused the notion of “prime” to be modified. In Z [ − 5 ] it can be proven that if any of the factors above can be represented as a product, e.g. 2 =

ab, then one ofaorbmust be a unit. This is the traditional definition of “prime”. It can also be proven that none of these factors obeys Euclid’s lemma; e.g. 2 divides neither (1 + √−5) nor (1 − √−5) even though it divides their product 6. In algebraic number theory 2 is calledirreduciblein Z [ − 5 ] (only divisible by itself or a unit) but notprimein Z [ − 5 ] (if it divides a product it must divide one of the factors). The mention of Z [ − 5 ] is required because 2 is prime and irreducible in Z . Using these definitions it can be proven that in any ring a prime must be irreducible. Euclid’s classical lemma can be rephrased as “in the ring of integers Z every irreducible is prime”. This is also true in Z [ i ] and Z [ ω ], but not in Z [ − 5 ].The rings in which factorization into irreducibles is essentially unique are called unique factorization domains. Important examples are polynomial rings over the integers or over a field, Euclidean domains and principal ideal domains.

In 1843 Kummer introduced the concept of ideal number, which was developed further by Dedekind (1876) into the modern theory of ideals, special subsets of rings. Multiplication is defined for ideals, and the rings in which they have unique factorization are called Dedekind domains.

Source: WikipediA

Fundamental Theorem of AlgebraThe

fundamental theorem of algebrastates that every non-constant single-variable polynomial with complexcoefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with an imaginary part equal to zero.Equivalently (by definition), the theorem states that the field of complex numbers is algebraically closed.

The theorem is also stated as follows: every non-zero, single-variable, degree

npolynomial with complex coefficients has, counted with multiplicity, exactlyncomplex roots. The equivalence of the two statements can be proven through the use of successive polynomial division.In spite of its name, there is no purely algebraic proof of the theorem, since any proof must use the completeness of the reals (or some other equivalent formulation of completeness), which is not an algebraic concept.

A first attempt at proving the theorem was made by d’Alembert in 1746, but his proof was incomplete. Among other problems, it assumed implicitly a theorem (now known as Puiseux’s theorem) which would not be proved until more than a century later, and furthermore the proof assumed the fundamental theorem of algebra. Other attempts were made by Euler (1749), de Foncenex (1759), Lagrange (1772), and Laplace (1795).

At the end of the 18th century, two new proofs were published which did not assume the existence of roots, but neither of which was complete. One of them, due to James Wood and mainly algebraic, was published in 1798 and it was totally ignored. Wood’s proof had an algebraic gap.

^{[3]}The other one was published by Gauss in 1799 and it was mainly geometric, but it had a topological gap, filled by Alexander Ostrowski in 1920, as discussed in Smale 1981 [3] (Smale writes, “…I wish to point out what an immense gap Gauss’ proof contained. It is a subtle point even today that a real algebraic plane curve cannot enter a disk without leaving. In fact even though Gauss redid this proof 50 years later, the gap remained. It was not until 1920 that Gauss’ proof was completed. In the reference Gauss, A. Ostrowski has a paper which does this and gives an excellent discussion of the problem as well…”).A rigorous proof was first published by Argand in 1806 (and revisited in 1813);

^{[4]}it was here that, for the first time, the fundamental theorem of algebra was stated for polynomials with complex coefficients, rather than just real coefficients. Gauss produced two other proofs in 1816 and another version of his original proof in 1849.None of the proofs mentioned so far is constructive. It was Weierstrass who raised for the first time, in the middle of the 19th century, the problem of finding a constructive proof of the fundamental theorem of algebra. He presented his solution, that amounts in modern terms to a combination of the Durand–Kerner method with the homotopy continuation principle, in 1891. Another proof of this kind was obtained by Hellmuth Kneser in 1940 and simplified by his son Martin Kneser in 1981.

Algebraic ProofsThese proofs use two facts about real numbers that require only a small amount of analysis (more precisely, the intermediate value theorem):

- every polynomial with odd degree and real coefficients has some real root;
- every non-negative real number has a square root.
The second fact, together with the quadratic formula, implies the theorem for real quadratic polynomials. In other words, algebraic proofs of the fundamental theorem actually show that if

Ris any real-closed field, then its extensionC=R(√−1) is algebraically closed.As mentioned above, it suffices to check the statement “every non-constant polynomial

p(z) with real coefficients has a complex root”. This statement can be proved by induction on the greatest non-negative integerksuch that 2^{k}divides the degreenofp(z). Letabe the coefficient ofzin^{n}p(z) and letFbe a splitting field ofp(z) overC; in other words, the fieldFcontainsCand there are elementsz_{1},z_{2}, …,zin_{n}Fsuch that

If

k= 0, thennis odd, and thereforep(z) has a real root. Now, suppose thatn= 2(with^{k}mmodd andk> 0) and that the theorem is already proved when the degree of the polynomial has the form 2^{k − 1}m′ withm′ odd. For a real numbert, define:

Then the coefficients of

q(_{t}z) are symmetric polynomials in thezs with real coefficients. Therefore, they can be expressed as polynomials with real coefficients in the elementary symmetric polynomials, that is, in −_{i}‘a_{1},a_{2}, …, (−1). So^{n}a_{n}q(_{t}z) has in factrealcoefficients. Furthermore, the degree ofq(_{t}z) isn(n− 1)/2 = 2^{k−1}m(n− 1), andm(n− 1) is an odd number. So, using the induction hypothesis,qhas at least one complex root; in other words,_{t}z+_{i}z+_{j}tzis complex for two distinct elements_{i}z_{j}iandjfrom {1, …,n}. Since there are more real numbers than pairs (i,j), one can find distinct real numberstandssuch thatz+_{i}z+_{j}tzand_{i}z_{j}z+_{i}z+_{j}szare complex (for the same_{i}z_{j}iandj). So, bothz+_{i}zand_{j}zare complex numbers. It is easy to check that every complex number has a complex square root, thus every complex polynomial of degree 2 has a complex root by the quadratic formula. It follows that_{i}z_{j}zand_{i}zare complex numbers, since they are roots of the quadratic polynomial_{j}z^{2}− (z+_{i}z)_{j}z+z._{i}z_{j}

Joseph Shipman showed in 2007 that the assumption that odd degree polynomials have roots is stronger than necessary; any field in which polynomials of prime degree have roots is algebraically closed (so “odd” can be replaced by “odd prime” and furthermore this holds for fields of all characteristics). For axiomatization of algebraically closed fields, this is the best possible, as there are counterexamples if a single prime is excluded. However, these counterexamples rely on −1 having a square root. If we take a field where −1 has no square root, and every polynomial of degree n ∈ I has a root, where I is any fixed infinite set of odd numbers, then every polynomial f(x) of odd degree has a root (since (x^{2}+ 1)^{k}f(x) has a root, where k is chosen so that deg(f) + 2k ∈ I). Mohsen Aliabadi generalized Shipman’s result for any field in 2013, proving that the sufficient condition for an arbitrary field (of any characteristic) to be algebraically closed is having a root for any polynomial of prime degree.^{[8]}

Another algebraic proof of the fundamental theorem can be given using Galois theory. It suffices to show thatChas no proper finite field extension.^{[9]}Let K/Cbe a finite extension. Since the normal closure of K overRstill has a finite degree overC(orR), we may assume without loss of generality that K is a normal extension ofR(hence it is a Galois extension, as every algebraic extension of a field of characteristic 0 is separable). Let G be the Galois group of this extension, and let H be a Sylow 2-subgroup of G, so that the order of H is a power of 2, and the index of H in G is odd. By the fundamental theorem of Galois theory, there exists a subextension L of K/Rsuch that Gal(K/L) = H. As [L:R] = [G:H] is odd, and there are no nonlinear irreducible real polynomials of odd degree, we must have L =R, thus [K:R] and [K:C] are powers of 2. Assuming by way of contradiction that [K:C] > 1, we conclude that the 2-group Gal(K/C) contains a subgroup of index 2, so there exists a subextension M ofCof degree 2. However,Chas no extension of degree 2, because every quadratic complex polynomial has a complex root, as mentioned above. This shows that [K:C] = 1, and therefore K =C, which completes the proof.Source: WikipediA

Another Proof## Here is the proof of the equivalent statement “Every complex non-constant polynomial p is surjective”.

1) Let C be the finite set of critical points , i.e. p′(z)=0 for all z∈C. C is finite by elementary algebra.## 2) Remove p(C) from the codomain and call the resulting open set B and remove from the domain its inverse image p−1(p(C)), and call the resulting open set A. Note that the inverse image is again finite.

## 3) Now you get an open map from A to B, which is also closed, because any polynomial is proper (inverse images of compact sets are compact). But B is connected and so p is surjective.

## I like this proof because you can try it for real polynomials and it breaks down at step 3) because if you remove a single point from the line you disconnect it, while you can remove a finite set from a plane leaving it connected.

## Source: mathoverflow

First Fundamental Theorem of CalculusLet

fbe a continuous real-valued function defined on a closed interval [a,b]. LetFbe the function defined, for allxin [a,b], by

Then,

Fis uniformly continuous on [a,b], differentiable on the open interval (a,b), and

for all

xin (a,b).Alternatively, if

fis merely Riemann integrable, thenFis continuous on [a,b] (but not necessarily differentiable).

CorollaryThe fundamental theorem is often employed to compute the definite integrals of a function

ffor which an antiderivativeFis known. Specifically, iffis a real-valued continuous function on [a,b], andFis an antiderivative offin [a,b], then

The corollary assumes continuity on the whole interval. This result is strengthened slightly in the following part of the theorem.

## Second Fundamental Theorem of Calculus

Let

fandFbe real-valued functions defined on a closed interval [a,b] such thatFis continuous on all [a,b] and the derivative ofFisffor almost all [a, b]. That is,fandFare functions such that for allxin (a,b), except for perhaps a countable number of points in the interval:

If

fis Riemann integrable on [a,b] then

The second part is somewhat stronger than the corollary because it does not assume that

fis continuous.When an antiderivative

Fexists, then there are infinitely many antiderivatives forf, obtained by adding an arbitrary constant toF. Also, by the first part of the theorem, antiderivatives offalways exist whenfis continuous.Source: WikipediA

## Fundamental Theorem of Galois Theory

Explicit Description of the CorrespondenceIn its most basic form, the theorem asserts that given a field extension

E/Fthat is finite and Galois, there is a one-to-one correspondence between its intermediate fields and subgroups of its Galois group. (Intermediate fields are fieldsKsatisfyingF⊆K⊆E; they are also calledsubextensionsofE/F.)For finite extensions, the correspondence can be described explicitly as follows.

- For any subgroup
Hof Gal(E/F), the corresponding fixed field, denotedE, is the set of those elements of^{H}Ewhich are fixed by every automorphism inH.- For any intermediate field
KofE/F, the corresponding subgroup is Aut(E/K), that is, the set of those automorphisms in Gal(E/F) which fix every element ofK.The fundamental theorem says that this correspondence is a one-to-one correspondence if (and only if)

E/Fis a Galois extension. For example, the topmost fieldEcorresponds to the trivial subgroup of Gal(E/F), and the base fieldFcorresponds to the whole group Gal(E/F).The notation Gal(

E/F) is only used for Galois extensions. IfE/Fis Galois, then Gal(E/F) = Aut(E/F). IfE/Fis not Galois, then the “correspondence” gives only an injective (but not surjective) map from { {\displaystyle \{} subgroups of Aut(E/F) } {\displaystyle \}} to { {\displaystyle \{} subfields ofE/F} {\displaystyle \}} , and a surjective (but not injective) map in the reverse direction. In particular, ifE/Fis not Galois, thenFis not the fixed field of any subgroup of Aut(E/F).## Properties of the Correspondence

The correspondence has the following useful properties.

- It is
inclusion-reversing. The inclusion of subgroupsHholds if and only if the inclusion of fields_{1}⊆ H_{2}E⊇^{H1}Eholds.^{H2}- Degrees of extensions are related to orders of groups, in a manner consistent with the inclusion-reversing property. Specifically, if
His a subgroup of Gal(E/F), then |H| = [E:E] and |Gal(^{H}E/F)/H| = [E:^{H}F].- The field
Eis a normal extension of^{H}F(or, equivalently, Galois extension, since any subextension of a separable extension is separable) if and only ifHis a normal subgroup of Gal(E/F). In this case, the restriction of the elements of Gal(E/F) toEinduces an isomorphism between Gal(^{H}E/^{H}F) and the quotient group Gal(E/F)/H.Source: WikipediA

## Fundamental theorem of Riemannian geometry

In Riemannian geometry, thefundamental theorem of Riemannian geometrystates that on any Riemannian manifold (or pseudo-Riemannian manifold) there is a unique torsion-free metric connection, called theLevi-Civita connectionof the given metric. Here ametric(orRiemannian) connection is a connection which preserves the metric tensor. More precisely:

Fundamental Theorem of Riemannian Geometry.Let (M,g) be a Riemannian manifold (or pseudo-Riemannian manifold). Then there is a unique connection ∇ which satisfies the following conditions:

- for any vector fields
X,Y,Zwe have

- ∂ X ⟨ Y , Z ⟩ = ⟨ ∇ X Y , Z ⟩ + ⟨ Y , ∇ X Z ⟩ , {\displaystyle \partial _{X}\langle Y,Z\rangle =\langle \nabla _{X}Y,Z\rangle +\langle Y,\nabla _{X}Z\rangle ,}
- where ∂ X ⟨ Y , Z ⟩ {\displaystyle \partial _{X}\langle Y,Z\rangle } denotes the derivative of the function ⟨ Y , Z ⟩ {\displaystyle \langle Y,Z\rangle } along vector field
X.

- for any vector fields
X,Y,

- ∇ X Y − ∇ Y X = [ X , Y ] , {\displaystyle \nabla _{X}Y-\nabla _{Y}X=[X,Y],}
- where [
X,Y] denotes the Lie bracket for vector fieldsX,Y.The first condition means that the metric tensor is preserved by parallel transport, while the second condition expresses the fact that the torsion of ∇ is zero.

An extension of the fundamental theorem states that given a pseudo-Riemannian manifold there is a unique connection preserving the metric tensor with any given vector-valued 2-form as its torsion. The difference between an arbitrary connection (with torsion) and the corresponding Levi-Civita connection is the contorsion tensor.

The following technical proof presents a formula for Christoffel symbols of the connection in a local coordinate system. For a given metric this set of equations can become rather complicated. There are quicker and simpler methods to obtain the Christoffel symbols for a given metric, e.g. using the action integral and the associated Euler-Lagrange equations.

Source: https://en.wikipedia.org/wiki/Fundamental_theorem_of_Riemannian_geometry

## The Fundamental Theorem of Linear Algebra

Not everyone knows about the fundamental theorem of linear algebra, but there is an excellent 1993 article by Gil Strang that describes its importance. For an

mxnmatrixA, the theorem relates the dimensions of the row space ofA(R(A)) and the nullspace ofA(N(A)). The result is that dim(R(A)) + dim(N(A)) =n.The theorem also describes four important subspaces and describes the geometry of

AandA^{t}when thought of as linear transformations. The theorem shows that some subspaces are orthogonal to others. (Strang actually combines four theorems into his statement of the Fundamental Theorem, including a theorem that motivates the statistical practice of ordinary least squares.)Source: WikipediA

## The Fundamental Theorem of Statistics

Although most statistical textbooks do not single out a result as THE fundamental theorem of statistics, I can think of two results that could make a claim to the title. These results are based in probability theory, so perhaps they are more aptly named fundamental theorems of probability.

- The Law of Large Numbers (LLN) provides the mathematical basis for understanding random events. The LLN says that if you repeat a trial many times, then the average of the observed values tend to be close to the expected value. (In general, the more trials you run, the better the estimates.) For example, you toss a fair die many times and compute the average of the numbers that appear. The average should converge to 3.5, which is the expected value of the roll because (1+2+3+4+5+6)/6 = 3.5. The same theorem ensures that about one-sixth of the faces are 1s, one-sixth are 2s, and so forth.
- The Central Limit theorem (CLT) states that the mean of a sample of size
nis approximately normally distributed whennis large. Perhaps more importantly, the CLT provides the mean and the standard deviation of the sampling distribution in terms of the sample size, the population mean μ, and the population variance σ^{2}. Specifically, the sampling distribution of the mean is approximately normally distributed with mean μ and standard deviation σ/sqrt(n).Of these, the Central Limit theorem gets my vote for being the Fundamental Theorem of Statistics. The LLN is important, but hardly surprising. It is the basis for frequentist statistics and assures us that large random samples tend to reflect the population. In contrast, the CLT is surprising because the sampling distribution of the mean is approximately normal

regardlessof the distribution of the original data! As a bonus, the CLT can be used computationally. It forms the basis for many statistical tests by estimating theaccuracyof a statistical estimate. Lastly, the CLT connects important concepts in statistics: means, variances, sample size, and accuracy of point estimates.Do you have a favorite “Fundamental Theorem”? Do you marvel at an applied theorem such as the fundamental theorem of linear programming or chuckle at a pseudo-theorems such as the fundamental theorem of software engineering? Share your thoughts in the comments.

https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem

## The Law of Large Numbers

Source: https://en.wikipedia.org/wiki/Law_of_large_numbers