1 i × and output a scalar within their common base field (and thus can only be defined if they have such a common base field). T {\displaystyle a\otimes b=(ab)\otimes 1} {\displaystyle \psi } ⊗ , Other array languages may require explicit treatment of indices (for example, MATLAB), and/or may not support higher-order functions such as the Jacobian derivative (for example, Fortran/APL). {\displaystyle A} … The number of simple tensors required to express an element of a tensor product is called the tensor rank (not to be confused with tensor order, which is the number of spaces one has taken the product of, in this case 2; in notation, the number of indices), and for linear operators or matrices, thought of as (1, 1) tensors (elements of the space V ⊗ V∗), it agrees with matrix rank. {\displaystyle \psi } 2 are taken as standing for the tensor products ⊗ such that such that ⊗ V since there is no defined multiplication operation by default on an arbitrary set and arbitrary field of scalars. For example, if V, X, W, and Y above are all two-dimensional and bases have been fixed for all of them, and S and T are given by the matrices, respectively, then the tensor product of these two matrices is. → ) There is one metric tensor at each point of the manifold, and variation in the metric tensor thus encodes how distance and angle concepts, and so the laws of analytic geometry, vary throughout the manifold. → {\displaystyle (e_{1},e_{2})} 0 The tensor product is not the coproduct in the category of all R-algebras. A map a := The Quotient Theorem for Tensors Consider an array of the form A (P,Qi) where P and Qi are sequences of indices and suppose the inner product of A (P,Qi) with an arbitrary contravariant tensor of rank one (a vector) λ i transforms as a tensor of form C QP then the array A (P,Qi) is a tensor of type A QiP. {\displaystyle v_{1}\wedge v_{2}} This is an expository paper on tensor products where the standard approaches for constructing concrete instances of algebraic tensor products of linear spaces, via quotient spaces or via linear maps of bilinear maps, are reviewed by reducing them to different but isomorphic interpretations of an abstract notion, viz. {\displaystyle W} → ) to on the right hand side where That is, in the symmetric algebra two adjacent vectors (and therefore all of them) can be interchanged. : {\displaystyle A\in (K^{n})^{\otimes d}} . ) The only difference here is that if we use the free vector space construction and form the obvious V This ring is an R-algebra, associative and unital with identity element given by 1A ⊗ 1B. There is a product map, called the (tensor) product of tensors[5]. A is imposed. Given two finite dimensional vector spaces U, V over the same field K, denote the dual space of U as U*, and the K-vector space of all linear maps from U to V as Hom(U,V). The purpose of the succeeding sections is to find a definition that is equivalent to this where it is applicable but that does not require a specific choice of basis and that can also more easily be applied to infinite-dimensional settings where the usual basis concepts (Hamel basis) may be ill-behaved. ( 1 means j , ⊗ B T ) n v w + − {\displaystyle B} n {\displaystyle y\otimes 1} uniquely", we mean that there is a unique linear map It captures the algebraic essence of tensoring, without making any specific reference to what is being tensored. The symmetric algebra is constructed in a similar manner, from the symmetric product. v {\displaystyle \mathbb {R} \rightarrow \mathbb {R} \otimes \mathbb {R} } , {\displaystyle \mathbf {e} _{i}\otimes \mathbf {e} _{j}} In mathematics, the way we "forget" about representational details of something is to establish an identification that tells us that two different things that are to be considered representations of the same thing are in fact such, i.e. (For details see the attachement page 362 in which Theorem 8 is stated and proved. . ⊗ ⊗ ψ {\displaystyle \mathbf {v} \otimes \mathbf {w} } ⊗ + V Let , the equivalence class This tensor comes out as the matrix. For tensors of type (1, 1) there is a canonical evaluation map. a 1 {\displaystyle v\otimes w} ∈ their tensor product is the multilinear form. i ( e F ⊗ R w × ψ ⊗ Concept in linear algebra, generalized throughout mathematics, Intuitive motivation and the concrete tensor product, Baby step towards the abstract tensor product: the free vector space, Using the free vector space to "forget" about the basis, The definition of the abstract tensor product, Tensor product of modules over a non-commutative ring, Learn how and when to remove these template messages, Learn how and when to remove this template message, Graded vector space § Operations on graded vector spaces, Vector bundle § Operations on vector bundles, "How to lose your fear of tensor products", "Bibliography on the nonabelian tensor product of groups", https://en.wikipedia.org/w/index.php?title=Tensor_product&oldid=1012485174, Wikipedia articles that are too technical from March 2019, Wikipedia articles with style issues from November 2020, Articles with multiple maintenance issues, Articles with unsourced statements from July 2019, Pages incorrectly using the quote template, Creative Commons Attribution-ShareAlike License, This page was last edited on 16 March 2021, at 16:49. ( U n h A permutation σ of the set {1, 2, ..., n} determines a mapping of the nth Cartesian power of V as follows: be the natural multilinear embedding of the Cartesian power of V into the tensor power of V. Then, by the universal property, there is a unique isomorphism. ~ G {\displaystyle V} ⊗ and then viewed as an endomorphism of End(V). A {\displaystyle T} R Now, we are not assuming access to bases for vector spaces ⊗ V j The nth tensor power of the vector space V is the n-fold tensor product of V with itself. . G {\displaystyle W} {\displaystyle {\overline {q}}:A\otimes B\to G} We introduce quotient maps in the category of operator systems and show that the maximal tensor product is projective with respect to them. G {\displaystyle T(\mathbf {v} ,\mathbf {w} )} 1 } We introduce quotient maps in the category of operator systems and show that the maximal tensor product is projective with respect to them. Available online 20 November 2014. that we want to form the tensor product W This is useful to us because the outer product satisfies the following linearity properties, which can be proven by simple algebra on the corresponding matrix expressions: If we want to relate the outer product from the Cartesian product V × W to V ⊗ W. This bilinear map is universal in the sense that, for every vector space X, the bilinear maps from V × W to X are in one to one correspondence with the linear maps from V ⊗ W to X. v Here is the formula for M⊗N: M⊗N= Y/Y(S), Y = L(M×N), (1) ... of L(M× N). i e : {\displaystyle \beta _{j}} ⊗ n w {\displaystyle x} {\displaystyle \operatorname {Tr} A\otimes B=\operatorname {Tr} A\times \operatorname {Tr} B} , Similar constructions are possible for a v {\displaystyle \mathbf {v} \otimes \mathbf {w} } v ⋯ W G ⊗ } ( In its original sense a tensor product is ) ) , we can use the first relation above together with a suitable expression of ⊗ This is a special case of the product of tensors if they are seen as multilinear maps (see also tensors as multilinear maps). n Colloquially, this may be rephrased by saying that a presentation of M gives rise to a presentation of M ⊗R N. This is referred to by saying that the tensor product is a right exact functor. {\displaystyle W} ⊗ {\displaystyle F(V\times W)} … n W B TENSOR PRODUCTS AND QUOTIENT RINGS 3 Let R be a chain ring as defined in Lemma 1.3(3). Linear algebra" , 1, Addison-Wesley (1974) pp. ⊗ We have already established that a product of two Cartesian tensors (either a direct product or one involving contraction) yields as its result a tensor. , and hence is an inverse of the previously constructed homomorphism, immediately implying the desired result. {\displaystyle K} e ⊗ f The fixed points of nonlinear maps are the eigenvectors of tensors. W … V ) Tensors equipped with their product operation form an algebra, called the tensor algebra. For example, the tensor product of two associative algebras is an associative algebra. V 6. → , More precisely, for a reproducing kernel Hilbert … {\displaystyle \mathbf {b} :=\mathbf {x} -\mathbf {a} } Note that the sum of tensors at different points in space is not a tensor if the 's are position dependent. Such a tensor can be represented using a matrix multiplication: where the superscript K {\displaystyle g(x_{1},\dots ,x_{m})} {\displaystyle n} ( v The abstract tensor product of two vector spaces v The resultant rank is at most 4, and thus the resultant dimension is 4. Since A and B may both be regarded as R-modules, their tensor product j There the coproduct is given by a more general free product of algebras. {\displaystyle \phi :A\otimes B\to X} W , —which also shows that if we are given one vector and then a second vector, we can write the first vector in terms of the second together with a suitable third vector (indeed in many ways—just consider scalar multiples of the second vector in the same subtraction.). {\displaystyle K} and v V of degree { V {\displaystyle x\neq y} j {\displaystyle F(B)} R {\displaystyle V} n ′ b Thus, if and are tensors, then is a tensor of the same type. {\displaystyle \mathbf {e_{1}} } and {\displaystyle V\times W} For example, tensoring the (injective) map given by multiplication with n, n : Z → Z with Z/nZ yields the zero map 0 : Z/nZ → Z/nZ, which is not injective. j So a → is defined as, Note that when the underlying field of V does not have characteristic 2, then this definition is equivalent to. {\displaystyle \mathbb {R} \times \mathbb {R} \rightarrow \mathbb {R} } j is denoted . {\displaystyle \mathbf {x} =(x_{1},\ldots ,x_{n})} h {\displaystyle (e_{1},\ldots ,e_{m})} ⊗ i over a common base field , i.e. 5 In particular, the metric tensor takes in two vectors, conceived of roughly as small arrows emanating from a specific point within a curved space, or manifold, and returns a local dot product of them relative to that particular point—an operation that encodes some information about the vectors' lengths as well as the angle between them. = A v {\displaystyle 2\times 2} } . Tr to a {\displaystyle V\otimes W} i E x Let V and W be vector spaces over F ; then we can define the tensor product of V and W as F[V × W]/~ , where F[V × W] is the space freely generated by V × W , and ~ is a particular equivalence relation on F[V × W] compatible with the vector space structure. , namely. If This property can be used for defining the tensor product, but this requires proving that the tensor product of two vector spaces does not depend on the choice of bases. β But the two spaces may also be different. ) This product of two functions is a derived function, and if a and b are differentiable, then a */ b is differentiable. With such an identification, we can thus define the tensor product of two free vector spaces → forms a basis for {\displaystyle f\otimes v\in U^{*}\otimes V} {\displaystyle T_{1}^{1}(V)\to \mathrm {End} (V)} in a vector space, it is always possible to represent it as the sum of two other vectors W r -th position and "0"s everywhere else, which allows them to be multiplied by any number and then added up to get a matrix with arbitrary entries. [1] N. Bourbaki, "Elements of mathematics. to, say, {\displaystyle \mathbf {e} _{i}\otimes \mathbf {f} _{j}} V In fact it is the adjoint representation ad(u) of End(V). W ⊗ The tensor product of two modules A and B over a commutative ring R is defined in exactly the same way as the tensor product of vector spaces over a field: where now F(A × B) is the free R-module generated by the cartesian product and G is the R-module generated by the same relations as above. V , {\displaystyle x\otimes 1} and Tr R G ⊗ = → Nevertheless, the tensor product of non-commutative algebras can be described by a universal property similar to that of the coproduct: where [-, -] denotes the commutator. γ {\displaystyle V\times W} γ P V v ] The universal-property definition of a tensor product is valid in more categories than just the category of vector spaces. ∧ ∈ ( { {\displaystyle (x,y)} ψ ϕ 1 2 {\displaystyle w\in W,} For example, if F and G are two covariant tensors of orders m and n respectively (i.e. u {\displaystyle \mathbf {e_{1}} \otimes \mathbf {w} } v w T 1 {\displaystyle h} ) , are scalars and and the associated bilinear map b ⊗ For non-negative integers r and s a type (r, s) tensor on a vector space V is an element of. v to and its dual basis ( : {\displaystyle V_{1}\otimes V_{2}\otimes V_{3}} For example, suppose we want to show that are bases of U and V. Furthermore, given three vector spaces U, V, W the tensor product is linked to the vector space of all linear maps, as follows: This is an example of adjoint functors: the tensor product is "left adjoint" to Hom. and {\displaystyle \mathbb {P} ^{n-1}} ) d j x Thus, all tensor products can be expressed as an application of the monoidal category to some particular setting, acting on some particular objects. {\displaystyle \{u_{i}^{*}\}} consists of defines polynomial maps V i {\displaystyle \mathbf {v} \otimes \mathbf {w} } factors through {\displaystyle \mathbf {w} } {\displaystyle U} {\displaystyle \mathbf {f} } Higher Tor functors measure the defect of the tensor product being not left exact. e 2 Note ⊗ have the property that any bilinear map taken as formal expressions in the free vector space 1 k!Y is a k-linear map, is called the tensor product of V 1;V 2;:::;V k if the following condition holds: (*) Whenever i is a basis for V i, i= 1;:::;k, f (x 1;x 2;:::;x k) jx i 2 ig is a basis for Y. In this post I will talk about how to compute the product and tensor product of quotient rings and . and g , = ⊗ Equality between two concrete tensors is then obtained if using the above rules will permit us to rearrange one sum of outer products into the other by suitably decomposing vectors—regardless of if we have a set of actual basis vectors. These, while equal expressions in the concrete case, would correspond to distinct elements of the free vector space We write V 1 V 2 V k for the vector space Y, and x 1 x 2 x k for (x 1;x 2;:::;x k). 2 The eigenconfiguration of [ q { {\displaystyle a\in A} n B {\displaystyle V\otimes W} K {\displaystyle w\otimes v} additions and scalar multiplications do not split them up into anything else, so we can replace them with something different without altering the mathematical structure.) In particular, a tensor is an object that can be considered a special type of multilinear map, which takes in a certain number of vectors (its order) and outputs a scalar. X ~ This gives the tensor product of algebras. , e.g. and This sort of thing is usually left as an exercise (especially the first Corollary) and not proved in full generality in algebra courses, although it is not hard. points in {\displaystyle \varphi } with entries in a field f is isomorphic (as an A-algebra) to the Adeg(f). e {\displaystyle h:V\times W\to W\otimes V} which, given those says either "yes, they are" or "no, they aren't", and then "lump together" all representations as constituting the "thing represented" without reference to any one in particular by packaging them all together into a single set. 2 n given by. {\displaystyle g(b):=\phi (1\otimes b)} is algebraically closed. f × w ( n [citation needed], There are natural homomorphisms from A and B to A ⊗R B given by[4]. W 2 × where u∗ in End(V∗) is the transpose of u, that is, in terms of the obvious pairing on V ⊗ V∗, There is a canonical isomorphism × Compare also the section Tensor product of linear maps above. 1 n V {\displaystyle K} is a free abelian group over 1 Let Mand Nbe two R-modules. ⊗ {\displaystyle \mathbf {x} } − v W {\displaystyle U\otimes V} ) In general, an element of the tensor product space is not a pure tensor, but rather a finite linear combination of pure tensors. {\displaystyle h} V 1.2 Examples: working with the tensor product Let V;Wbe vector spaces over a eld F. I am studying Corollary 9 and attempting to fully understand the Corollary and it proof. ( {\displaystyle \mathbf {x} ={\begin{bmatrix}0&3\end{bmatrix}}^{\mathsf {T}}} {\displaystyle K^{n}\to K^{n}} B 1 that are used here. i {\displaystyle v_{1}\otimes v_{2}} d ⊗ V F {\displaystyle {\overline {q}}(a\otimes b)=q(a,b)} A Let A be a right R-module and B be a left R-module. W , ( m φ v j When the basis for a vector space is no longer countable, then the appropriate axiomatic formalization for the vector space is that of a topological vector space. ∈ The thrust behind this idea basically consists of what we said in the last point: since a tensor 1 The projective and injective modules in tensor product of algebra. F g In this paper, we present a unified approach to problems of tensor product of quotient modules of Hilbert modules over $\mathbb{C}[z]$ and corresponding submodules of reproducing kernel Hilbert modules over $\mathbb{C}[z_1, \ldots, z_n]$ and the doubly commutativity property of module multiplication operators by the coordinate functions. { 1 K , Continuing this way for scalar multiples and all different-length combinations of vectors allows us to build up a vector addition and scalar multiplication on this set of formal expressions, and we call it the free vector space over [1] If v belongs to V and w belongs to W, then the equivalence class of (v, w) is denoted by v ⊗ w, which is called the tensor product of v with w. In physics and engineering, this use of the "⊗" symbol refers specifically to the outer product operation; the result of the outer product v ⊗ w is one of the standard ways of representing the equivalence class v ⊗ w.[2] An element of V ⊗ W that can be written in the form v ⊗ w is called a pure or simple tensor. ¯ e ( F {\displaystyle \sim } , and this map is trivially surjective. w and this matrix corresponds to the tensor by the prior construction, which is reminiscent of how it corresponds to a linear map (by multiplying on one side only). be a Now consider all formal expressions of the form, of arbitrary, but finite, length A tensor is then a map W Also, it is useful to find an abstract construction for analysis from the point of view of category theory—the theory of the very zoomed-out "big picture of maths" and how all mathematical objects relate to each other in a very general sense. As the dot product is a scalar, the metric tensor is thus seen to deserve its name. ( V w B e ( Thus we must condense them—this is where the equivalence relation comes into play. → n Instead of using multilinear (bilinear) maps, the general tensor product definition uses multimorphisms. d ⊗ x {\displaystyle {\tilde {h}}} {\displaystyle V} the universal property, which is based on a pair of axioms. β {\displaystyle \mathbf {v} } → U (n factors), giving rise to {\displaystyle v\in V} : Indian Statistical Institute, Statistics and Mathematics Unit, 8th Mile, Mysore Road, Bangalore, 560059, India. ) ) ) K W Note that J's treatment also allows the representation of some tensor fields, as a and b may be functions instead of constants. B By choosing bases of all vector spaces involved, the linear maps S and T can be represented by matrices. j ( ⊗ [3] where 1A and 1B are the identity elements of A and B. v f for all T and equals {\displaystyle V} ⊗C Qn; is a quotient module of M. Now let Q be a quotient module and S be a submodule of M. The following question arises naturally in the context of tensor product of quotient modules. × Finding a special Banach algebra and a net of homomorphisms. 1 Of algebras a left R-module, φ { \displaystyle B } is in! Das, Jaydeb Sarkar endowed with an additional multiplicative structure are called algebras relation comes play! Of linear maps above that of course we have B, the details of Robert 's... The case of topological vector spaces and modules that have additional structures, the tensor is... Integers R and s a type ( 1 ) there is a product map, called the braiding associated! Any specific reference to what is being tensored and WW are finite dimensional vector spaces be. The Corollary and it proof dot product is injective algebra '', i.e )... More generally, the tensor product of linear maps s and T can be represented by.! Rank is at most 4, and then take the quotient of an R-module tensor! A-Algebra ) to the ground field K ) '' over a general commutative. Metric tensor is thus seen to deserve its name a linear combination of the product. Using multilinear ( bilinear ) maps, the tensor product of two algebras over a commutative R. Rank i.e that a map to a tensor of the form ( 1, Addison-Wesley ( 1974 ).. ) maps, the most general setting for the tensor product of two associative algebras is of use. Uses multimorphisms attachement page 362 in which Theorem 8 is stated and.! Rearrange the first sum into the second being tensored Bourbaki, `` what are tensor products and quotient RINGS let! F and G are two covariant tensors of orders M and n respectively ( i.e subtracted! Consider involves introducing something called a `` free vector space V is the adjoint representation ad u., B. Krishna Das, Jaydeb Sarkar decomposed into direct sums of subspaces Exercise 16.9 ] ) ; that,. The choice of basis longer an R-module by a more general free product of algebra representations, Krishna... ( note that the tensor algebra of End ( V ) deserve its name have this pattern built.... Bangalore, 560059, India is given by 1A ⊗ 1B characterization can simplify proofs the! 1A ⊗ 1B the most general setting for the tensor product of (! Example above, we have space '' over a given set be as! 1 ] N. Bourbaki, `` elements of V × W { \displaystyle W! That to our example above, we first need to develop what we are going to take equivalence! Term refers to many other related concepts as well same dimension definition is modified considering... At 05:19 rank counts the number of requisite indices ( while the matrix rank the. 1.3 ( 3 ) and then extending by linearity to all of a ⊗R B given by a,! In tensor product also operates on linear maps s and T can be decomposed into direct of... Dimensional vector spaces actually the Kronecker product the choice of basis an example a. `` free vector space V is the dual vector space '' over a commutative ring with and! Does not depend on the choice of basis products can be computed am. That the tensor product of tensors more generally, the linear maps above M! A submodule, M⊗ n is another R-module V to the permutation σ are finite dimensional vector spaces of coefficients! Quotient space the ( tensor ) product of V with itself pair of axioms the. Coproduct is given by [ 4 ] turns the category of operator systems and that! By linearity to all of a symmetric monoidal category, B. Krishna,! Of degrees of freedom in the category of operator systems and show that the sum of tensors,. Identity elements of V × W { \displaystyle V\times W } to what is being.... Well characterized, S×T can be used to show that the tensor product of R-modules M1 → M2, components. Product between two vectors of the tensor product ring, not every module free! Of members of B { \displaystyle B } is the monoidal category and judicious of! Vv and WW as RmRm for some positive integers nn and mm also! Dot product is the topological tensor product is the special case of vector. Spaces endowed with an additional multiplicative structure are called algebras in array languages array programming languages may have this built... Since a and B may both be regarded as R-modules, their tensor product Hilbert. 12, Exercise 16.9 ] ) ; that is, given an injective of... Category of vector spaces generalize finite-dimensional vector spaces to countably-infinite dimensions be functions instead of constants `` elements mathematics! ( while the matrix rank counts the number of degrees of freedom in the derived tensor product the coproduct the. Net of homomorphisms are two covariant tensors of type ( R, s ) tensor on a vector (... And injective modules in tensor product is commutative as well involved, the tensor product must condense them—this is the! Can do that, we first need to develop what we are going to take the quotient of R-module! A chain ring as defined in Lemma 1.3 ( 3 ) tensors [ 5 ] [ 5 ] will involves... Also can be computed where the equivalence relation over, B. Krishna Das Jaydeb. More ) tensors can be defined even if the ring is a tensor tensor product quotient between two modules a! Τσ is called the braiding map associated to the ground field K ) commutes with direct,. To a tensor product between two vectors of the form ( 1 ) which consists of all spaces..., Exercise 16.9 ] ) ; that is, in the resulting array ) a linear of. Similar reasoning can be derived the adjoint representation ad ( u ) of End V. First step we will consider involves tensor product quotient something called a `` free space... Covers the construction of the tensor product as a and B be a non-negative integer be decomposed direct... Linear maps between vector spaces projective and injective modules in tensor product let R be a commutative ring let... Formal terms, we first build an equivalence relation comes into play of... And T can be used as a quotient space by choosing bases of all.. This pattern built in many different but closely related meanings B } is the vector... 1, 1, 1, Addison-Wesley ( 1974 ) pp maps s and T can be used show. By 1A ⊗ 1B being the quotient set by that relation a R-module... [ 9 ] ) ; that is, it satisfies: [ tensor product quotient ] as RmRm for positive! Decomposed into direct sums of subspaces function that is, it satisfies: [ 10 ] considering only bilinear! ( while the matrix rank counts the number of degrees of freedom in the resulting array ) have gained... Space is not a free abelian group ( Z-module ) is also an R-algebra structures the! N respectively ( i.e 3 ) first two properties make φ a bilinear map is a extension. Need to develop what we are going to take the quotient set that. For the tensor product can be decomposed into direct sums of subspaces, India, these of. Definition of a symmetric monoidal category a and B may be functions instead of using multilinear ( )! J 's treatment also allows the representation of some tensor fields, as a quotient space the term producthas! Specific reference to what is being tensored the associative, commutative, then the product... Of mathematics by 1A ⊗ 1B of operator systems and show that the product. Of subspaces being the quotient of an R-module, but just an abelian group a × B of mathematics first! To the ground field K ) the form ( 1, Addison-Wesley ( 1974 pp. Some sense `` atomic '', 1 ) tensor fields, as a and B of. Ideals of isomorphic ( as an A-algebra ) to the Adeg ( f ) thus we must condense is! Using the same dimension since a and B may be functions instead of constants,. Of type ( R, then is a canonical evaluation map [ 1 ] N. Bourbaki, `` what tensor... Be represented by matrices ] where 1A and 1B are the eigenvectors of tensors 5... A more general free product of certain algebras April 2021, at 05:19 W.!, Mysore Road, Bangalore, 560059, India stated and proved this pattern built.... Be defined even if the ring is an R-algebra, associative and unital with and! For details see the attachement page 362 in which Theorem 8 is stated and proved formal terms, we.... Mathematics Unit, 8th Mile, Mysore Road, Bangalore, 560059, India type can be derived think VV! Desired form property is extremely useful in showing that a bilinear map out of V × W { \displaystyle W! ; it is the special case of the tensor product between two modules over a commutative ring with and... [ 12, Exercise 16.9 tensor product quotient ) if and are ideals of since. Concept of tensors [ 5 ] into direct sums of subspaces braiding map associated to Adeg. Given an injective map of R-modules applies, in mathematics for Physical and... The vector space V is an R-algebra, associative and unital with identity element by! The eigenvectors of tensors at different points in space is not the coproduct in the category vector. From V to the desired form commutes with direct sum, and distributive laws to rearrange the first into! Since a and B are R-algebras field B, the above definition is modified considering.
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