Wait! You mean to say that the floor here is made of floor?!

@@Erik_Caballero precisely that

I mean, a "vector" in mathematics is "an element of a vector space", so that tensor definition seems perfect to me.

@@theadamabrams And let's not forget that a "vector space" in turn is simply a "space consisting of vectors" xD

my friend used to always say aloud when heard this definition "ok boomer" :D

Hey Andrew so cool to see you here! You should cover this in a drink and derive video ;)

ayyy smart people

Wait. Wasn't a tensor just something that transforms like a tensor? We have been lied!

@@axelperezmachado3500 physicists have subtly different definitions for vectors and tensors than mathematicians do.

By the way i loved the video on motivation

Hey if spamming videos in the comments isn't allowed, I'm happy to not do it in the future, but I just made a short video about this on my channel.

@@k-theory8604 if it is related to the topic at hand, it’s not really shameless advertisement and we appreciate the reference. In this case it’s fine.

Tensors as defined in this video are objects defined over a vector space. Tensors used in physics are really tensor fields on a manifold M. For every point p on a manifold, we have a natural vector space given by the tangent space TpM. On each tangent space, we can define some sort of tensor. A physicist's tensor is a smoothly varying field of tensors. A nice fact about manifolds is that they can be locally parameterized by Euclidean space. These parameterizations are usually called charts. We can use charts to parameterize all tangent spaces within a neighborhood of any given point. This allows us to write down the components of a smoothly varying field of tensors using the coordinates given by the chart. Then comes the "tensor transformation law." If some point p sits in the intersection of two different charts, we have two sets of local coordinates coming from each chart. The tensor transformation law simply says that changing from one set of coordinates A to the other set of coordinates B should change the components of the tensor in such a way that the local description of the tensor in the B coordinates matches the transformed version of the tensor in the A coordinates. You can think about this compatability condition as gluing the local descriptions of the tensor together to form a global tensor field on the whole manifold. The concepts are related but not identical. To summarize: a mathematician's tensor is an object defined over a vector space, a physicist's tensor is a field of mathematician's tensors on a manifold.

Its the same thing, this is just the definition.

It is related by being the same fucking concept.

I second that. Maybe with implication of a base change for the vectors (like cartesian to oblique, or polar (2D), or curved). Sure that will be even greater with derivatives (first and second order).

lol I thought the most advanced math engineers used was basic math like vector calc and linear algebra and some integral transforms

@@mastershooter64 it depends where you study :-)

@@cicciocareri85 which field of engineering uses the most advanced math?

@@mastershooter64 Finite elements technique often use simple "reference elements" which are then ... tortured. Depending of the desired "continuity", the numerical derivatives implied by the (partial) differential equation(s) borrow heavily from tensor theory.

My understanding is that the tensors in physics are just these simple tensors that Michael is talking about, but using its matrix representation instead. But just like you can talk about a whole matrix (a_ij) by only considering a single general entry a_ij, physicists prefer to only talk about a single entry in their tensors, which may also require superscripts.

I think tensors in physics, at least in general relativity, are really tensor fields from differential geometry.

​@@JM-us3fr AFAIK in physics there's an explicit restriction with how tensors transform under coordinate transformations. Is this restriction also implicit in the mathematical definition, or no?

@@rtg_onefourtwoeightfiveseven I’m not sure if these are the same, but I know when you want to change the basis of your vector space, and you want to represent an alternating tensor with respect to the new basis, then you can just multiply the 1st representation by the determinant of the basis transformation matrix. But you’re probably referring to the reference frame transformation in general relativity. I’m not quite sure how this works.

@@rtg_onefourtwoeightfiveseven Yes it is implicit in the mathematical definition. You start with a differentiable manifold which has local coordinate charts, but you define tensor fields on the manifold independently of coordinate charts. So, they don't depend on the choice of coordinates. Let us work point-wise on the manifold. When you see the tensors in GR, what you are looking at is a choice of coordinate chart which determines a basis for the tangent/cotangent spaces (these are vector spaces associated to points on the manifold), which then allows you to express tensors as linear combinations of tensor products of these basis vectors. The coefficients of these linear combinations are the coefficients you see in Einstein notation in GR; these are, mathematically, NOT tensors. Now, I said the tensors themselves are completely coordinate independent. What this means is that the basis vectors (determined by a coordinate transformation) will transform inversely to the coefficients so that they cancel out. This is where the Jacobian/inverse Jacobian come in for the coordinate transformations. Finally, the upper and lower indices of tensors really refer to the tangent and cotangent spaces, where the cotangent space is the dual of the tangent space. A rank (1,1) tensor is built from a linear of combination of tangent vectors tensor producted with cotangent vectors. Now, if the manifold has a Riemannian metric, then it can be used to provide a canonical isomorphism between tangent and cotangent spaces (given by raising/lowering indices with the metric in GR notation). Hopefully this gives some ideas. There is quite a lot going on and I have swept some things under the rug, but if you want to formalise this further you need: vector bundles, tensors products of vector bundles, sections of bundles, and universal property of tensor products

Yeah, I thought this was very important too

The way I understand it, the span of a set is the set of all finite linear combinations of elements of that set, and the formal product is infinite dimensional because any v*w is a basis vector, since it cannot be expressed as a linear combination of other elements of that type.

The dimension is the maximum number of independent vectors you can have. R^3 has at most 3: for example, [1,0,-9], [0,4,0], [1,2,3]. You put any more vectors, and you get relations between them. For the star product, every v*w is independent from every r*s (where either component is different). You have infinitely many options for the first component, then infinitely many options for the second, so the dimension is infinite.

Is the formal product essentially just the Cartesian product or is there a difference I’m missing?

@@brooksbryant2478 The basis of the star product is just the Cartesian product. But the span is much bigger. You're taking every (finite) linear combination from this basis to make the star product space.

Formal product is basically a “dummy” operation. You basically just stick the events together, but you are not allowed to do anything to them. So, for example, if * is a formal product, you could NOT say that 2*3=3*2 or anything like that. 2*3 and 3*2 are their own separate things.

The short version is, the "tensors" (letters with upper and lower indices) that appear in physics (at least in general relativity; I know little about, say, material stresses) are the coefficients of the linear combinations we see here. The basis vectors and the tensor product symbol are elided.

​@@MasterHigure No, the tensors are not the coefficients. In physics, we are often concerned with vector fields. Essentially, we imagine these as some sort of vector sitting at each point in space, which vary continuously. However, in order to formally define such a thing, one first needs to define the tangents space to a (manifold) space at every point. Then, one thinks about all of these tangent spaces together as forming one large geometric object, called the tangent bundle Even more generally, one takes the tensor products of each tangent space with itself a number of times, and then consider that as one big collection of objects. Then, tensor fields in physics are just maps from the underlying space to the (generalized) tangent bundle. TL;DR: They are related, but the physics tensor is much more complicated. One needs the definition in this video to define the version from physics.

Yeah exactly, then you also need the cotangent bundle and the notion of the universal property of tensor products. Then, after specifying a coordinate chart, you can evaluate everything using the basis provided by the chart and get the usual physics Einstein notation. Quite a journey lol

Don't fear my Levi-Civita operator and my Electromagnetic tensor

@@thenateman27 Electromagnetic 2-form*

What is the star operation between vectors??

What is the star operation between vectors?? Please develop

The small star between two vectors is a 'dummy' Operation. So * Maps v and w to v * w wich is an Element in V * W. Therefore the Star is just notation. Then he says all these star products are defined to be linear Independent. Hence all Elements v * w are a Basis of V * W. (This yields for example immediatly, that V*W is Infinite dimensional). I hope this helps you. Edit: an example: in the karthesian product (a, b) + (c, d) = (a+c, b+d), however a*b + c*d is not equal to (a+c)*(b+d).

@@philippg6023 So, if I get you correctly, there is no "star" operation per se but it's a representation of any linear operation that has certain characteristics. Is that correct?

@@Fetrovsky maybe you mean the correct Thing but you say it a little bit wrong. v * w is Just a symbol, which has No meaning at all. Hence the Star product is called formal product If you have a good understanding of Basis, then you might Like this explenation. V*W is a vectorspace with the karthesian product as Basis.

@rachmondhoward2125 Insanely beautiful construction example. Let's (A)(D)simplicyan be positional natural A-ary SUM(D)-digit number system. where A,D -some natural number sequences arbitrary, but equal length. Then: d-vertex simplex is a (2)(d)simplicyan. square is a (3)(2)simplicyan. cube is a (3)(3)simplicyan. d-dimensional (n-1)*2 -gon is a (n)(d)simplicuan pentagon is a (11)(1)simplicyan. d-dimensional (n-1)/2 -gon is a (n)(d-1)simplicuan Maya/Egypt pyramid is a (2,3)(1,2)simplicyan, let's enumerate it: 100 is a vertex above square. 022 (=NOT 100) is opposite edge 020 002 is are 2vertices between wich lies edge 022=002+020. 120 (=100+020) is edge between verices 100 and 020. 102 (=100+002) is edge between verices 100 and 002. 122 (100+020+002) is infinity(content) located between 3 vertices 100 020 002 and 3 edges 120 102 022. from that moment we have 2 variants: a)vertex 001 connected to vertex 020.(i choose it) b)vertex 001 connected to vertex 002. 001 (=100+100) is vertex connected to vertex 100. 021 (=001+020) is edge between vertices 001 and 020. 010 (=001+001) is vertex connected to vertex 001. 011 (=001+010) is edge between vertices 001 and 010. 012 (=002+010) is edge between vertices 002 and 010. 101 (=001+100) is edge between vertices 001 and 100. 110 (=100+010) is edge between vertices 100 and 010. 111 (=100+010+001) is triangle between vertices 100 010 001. 121 (=100+020+001) is triangle between vertices 100 020 001. 112 (=100+010+002) is triangle between vertices 100 010 002. 000 is zero is externity of pyramid . As you see: 1. infinity(content) is like triangle. 2. square doesn't exists, sguare + volume above it upto vertex 100 is Zero.

@@user-ys3ev5sh3w Your construction has numerous mistakes and inconsistencies: 1. You count interior of a k-simplex, but you don't count interiors of a square and cube. 2. Compared with (n-1)/2-gon, your (n-1)*2-gon has 4 times more "gon"s (whatever they are), but its corresponding simplician has n times more indices. 3. Equation "001=100+100" is false. 4. And what about dodecahedron? It's 3-dimensional 12-hedron with 30 edges, but doesn't correspond neither to (7)(3) nor (16)(3) simplician.

@@Mr.Not_Sure you are right. In equation "001=100+100" i mean that carry rotate. Preface. Positional natural a-ary (az^m) d-digit number systems can represent some kind of d-dimensional polytopes in geometry. If a=z^m then a-ary d-digit number system equals z-ary (m*d)-digit number system, because z^m^d=z^(m*d). For example: binary d-digit number system is a d-vertex simplex.(vertices is a numbers with digital root=1, edges is a numbers with digital root=2 and so on). 3-ary d-digit number system is a d-cube. Let d=2: if 1-chain (2 vertices + 1 edge) shift 1 times we receive 2-cube with 3^2=9 faces, i.e. square. 6-ary d-digit number system is a d-thorus. Let d=2: If 3-ring (1D triangle) shift in 3D 3 times and "press"(glue) first and last 1D triangles we recieve 2-thorus with 6^2=36 faces = 3*3 square + (3+3)*3 edges + 3*3 vertices. Conclusion. All positional natural a-ary (az^m) d-digit number systems with a^d numbers are represented by 3 types of d-polytopes (simplex (binary), cube (odd-ary) , thorus (even-ary) ) with a^d faces . For simplex d means amount of vertex or hyperplanes, for any other polytopes dimension. For me as a programmer, it's curious to know that difference in faces between consequent such polytopes is hexagonal numbers. To enumerate more complex polytope may be used positional natural A-ary D-digit number system , (A)(D)simplician. Where A,D - some natural number sequence arbitrary but equal length. Then 2D Maya/Egypt pyramid is are positional natural (2,3^2)-ary (1,1)-digit number system or (2,3^2)(1,1)simplician. digital root of (1,1)=2 is dimension of pyramid. 10 is a vertex above square. 08 (=NOT 10) is one of opposite edges. 06 02 is are 2 vertices between which lies edge 08=02+06. 16 (=10+06) is edge between vertices 10 and 06. 12 (=10+02) is edge between vertices 10 and 02. 18 (10+06+02) is triangle, infinity(content) located between 3 vertices 10 06 02 and 3 edges 16 12 08. 01 (=10+10) is vertex connected to vertex 10. 07 (=01+06) is edge between vertices 01 and 06. 03 (=01+opposite vertex 02) is vertex connected to vertex 01 and 02 instead of absent edge between 01 and 02.(exeption from rule?). 04 (=01+03) is edge between vertices 01 and 03. 05 (=02+03) is edge between vertices 02 and 03. 11 (=01+10) is edge between vertices 01 and 10. 13 (=10+03) is edge between vertices 10 and 03. 14 (=10+03+01) is triangle between vertices 10 03 01. 17 (=10+06+01) is triangle between vertices 10 06 01. 15 (=10+03+02) is triangle between vertices 10 03 02. 00 is square(zero). 18 faces=5 vertices+8 edges+4 triangles+1 square=3^2*2^1 numbers. 2D dodecahedron has 12 pentagons+30 edges+20 vertices=62 faces= 31^1*2^1 numbers of (2,31)(1,1)simplician (even thorus-like 2-polytope). 3D dodecahedron has 62+I=63 faces =7*3*3 numbers of (7,3)(1,2)simplician (odd cube-like 3-polytope). Simplex differs from any other polytope. Simplex is only "not preesed" polytope. Outside and Inside parts of this polytope is considered by me to be faces. As result simplex has 2^d faces (no 2^d-1). If to increment dimension they became from potentional to real faces. Outside and Inside parts both are counted. In other polytopes 1 or 1/2 of dimension is "pressed"("curved") in variouse ways. For example in even(thorus-like) 2-dimensional in 3D (no 3-dimensional) M aya/Egypt pyramid Outside part of "pressed" dimension is square, Inside part is triangle, so both Outside and Inside parts (interior and exterior) are not counted. In odd (cube-like, in A odd numbers only) polytope only Outside part (zero) is pressed in one of hyperplanes, so Outside part is not counted, only interior is counted, . And so on. "No distinction between numbers and shape. Numbers could not exist without shape." Pythagoras (reincarnation of Euphorbos).

@@Mr.Not_Sure Lets enumerate 3D dodecahedron ,wich is (7,3)(1,2)simplician with 7^1*3^2=63 faces. It's evident: 1. (m+n,..)(1,..)=(m,..)(1,..)+(n,..)(1,..) because of (m+n)^1=m^1+n^1. 2. (m,..)(i+j,..)=(m,..)(i,..)*(m,..)(j,..) because of m^(i+j)=m^i*m^j. 3. (m,m,..)(i,j,..)=(m,..)(i+j,..) because of m^i*m^j=m^(i+j). 4. (m^n,..)(i,..)=(m,..)(n*i,..) because of m^n^i=m^(n*i). 5. (m,n,..)(i,i,..)=(m*n,..)(i,..) because of m^i*n^i=(m*n)^i. 6. ( m ,1,..)(i,j,..)=(m..)(i,..) because of m^i*1^j=m^i.. Use 1..4: (7,3)(1,2)=(4+3,3)(1,2)=(4,3)(1,2)+(3,3)(1,2)=(2,3)(2,2)+(3)(3)=(6)(2)+(3)(3); So 3D dodecahedron equals 3D cube (3)(3) + 2D thor (6)(2). Lets enumerate 1D chain (3)(1). It's easy. Let edge be 2 because 2 is maximal number(infinity) Let rest two vertices be 0 1. Lets enumerate 2D square (3)(2)=(3)(1)*(3)(1). It's easy. Let 2D chain in center be located in 2 of first digit. Let rest two 1D chains be located in 0 1 of first digit. Lets enumerate 3D cube (3)(3)=(3)(1)*(3)(2). It's easy. Let 3D chain in center be located in 2 of first digit. Let rest two 2D chains be located in 0 1 of first digit. Lets enumerate 1D ring(triangle) (6)(1)=(3)(1)+(3)(1). It's easy. Let 1D chain be located in 0..2 of digit Let rest inversed 1D chain located in 3..5 of digit. Lets enumerate 2D thor (6)(2)=(6)(1)*(6)(1). It's easy. Let 3 1D rings be located in 0..2 of first digit Let 3 2D rings be located in 3..5 of first digit. Lets enumerate 3D dodecahedron (7,3)(1,2)=(6)(2)+(3)(3). It's easy. Let 3D cube (3)(3) be located in 0..2 of first digit unmodified as above. Let 2D thor (6)(2) be located in 3..6 of first digit in such way: (6)(2)=(4,3)(1,2)=(2,3)(1,2)+(2,3)(1,2)= 2D maya/egypt pyramid +2D maya/egypt pyramid. Let 2D maya/egypt pyramid be located in 3..4 of first digit Let 2D maya/egypt pyramid be located in 5..6 of first digit At last we must "press" 6 vertex of cube to triangles. As result 6 vertex of cube became interiors of 6 hedrons. 6 rings of thor are added to close this 6 new interiors. We receive 12 hedrons=6 old + 6 new. 30 edges=12 edges of cube + 3*3 edges + 3*3 square of thor 20 vertices=(8-6) vertices of cube + 3*3 vertices of 3 1D rings +3*3 edges of 3 2D rings of tohr.

I think getting desired properties via a quotient is a fairly general technique, isn't it? That's also how you can get complex numbers out of pairs of real numbers. Or how you can get interesting groups out of free groups etc.

Read Spivak's book even now. It is the most intuitive (to me at least), abstract exposition of tensors, chains, differential forms, etc I could find.

I think the word "tensor" in machine learning is closer to the programming notion of "array": The objects described in this video have more structure, in some sense.

“More structured” or “more constrained”?

@@AdrianBoyko I think I meant what I said. Tensors in math are associated with the notion of multilinear functions, but in a neural network that does image recognition, the input can be a "tensor" with dimensions width x height x 3(rgb) x minibatch_size. However, as far as I can tell, that's just an array, without any further structure.

@@alonamaloh many of the operations done with them are multilinear maps. Specifically, the parts with the parameters that are trained are often multilinear maps with these, with non-linear parts between such multilinear maps. This seems to me somewhat tensor-y ? Like, yeah, it is all done with a basis chosen for each vector space, but, well, you *can* reason about it in a basis-free way ? Like, for an image with 3 color channels, the space of such images can be regarded as the tensor product of the space of greyscale images with the space of colors, or as the tensor product of greyscale 1D horizontal images with greyscale 1D vertical images with colors. The space of translation-invariant linear maps from greyscale images to greyscale images (of the same size) (so, convolutions) is a vector space, and I think it is the tensor product of the analogous spaces for the spaces of the horizontal and vertical 1D images. This all seems like tensors to me?