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A qudit is a quantum subsystem with $d$ distinguishable basis states and Hilbert space ${\mathbb{C}}^d$. A qubit is the special case $d=2$. Tensor products of qudits give a natural setting for local Hamiltonians and finite-dimensional quantum circuits.

The closed-string sector with Ramond sector states on both chiral sides. The left and right spinor tensor product describes differential-form gauge fields. Opposite chiralities in type IIA superstring theory give the massless one-form and three-form, with transverse dimensions $8$ and $56$. Same chiralities instead give the type IIB form-field content.

The rational homotopy group is obtained by tensoring a homotopy group with the rational numbers. It retains the free part and discards torsion.

For simply connected $X$, rationalization of a topological space is a map $r:X\to X_{\mathbb{Q}}$ to a rational space inducing ${\pi }_n(X_{\mathbb{Q}})={\pi }_n(X)\otimes \mathbb{Q}$. It is also a rational homology equivalence. Its universal property is the bijection $r^{\ast }:[X_{\mathbb{Q}},Z]\to [X,Z]$ for every rational space $Z$. A Postnikov tower constructs it by replacing each fibre group with its rational tensor product and transporting the Postnikov invariant through the resulting cohomology isomorphism, then taking the homotopy inverse limit.

The representation ring is the Grothendieck group of finite-dimensional complex unitary representations of a compact group, with addition induced by direct sum and multiplication by tensor product. As an abelian group it is freely generated by the irreducible unitary representations. Taking dimension gives a ring homomorphism $R(G)\to \mathbb{Z}$.

Tensor products $V_{j_L}\otimes V_{j_R}$ give the irreducible representations of the Spin(4) double cover. The simultaneous central sign acts by $(-1{)}^{2j_L+2j_R}$, so exactly the integer-sum pairs descend to SO(4). Their dimensions are $(2j_L+1)(2j_R+1)$. Under the diagonal spatial rotation subgroup, the Clebsch-Gordan decomposition for SU2 gives spins from $|j_L-j_R|$ through $j_L+j_R$ in steps of one.

Consider a mechanically free isotropic model of a saturated ferromagnet, at fixed temperature and fixed magnitude of magnetization. Let $\mathbf{m}=\mathbf{M}/M_s$ be a unit vector, where $M_s$ is the saturation magnetization. The magnetostrictive strain is an infinitesimal strain tensor, measured relative to a reference with randomly oriented magnetic domains. This reference convention matters: it sets the orientational average of the preferred strain to zero. Define ${\lambda }_s$, the saturation magnetostriction, as the longitudinal relative length change in this model.

Rotational symmetry permits a symmetric second-rank tensor made from $\mathbf{m}$ of the form $a{\delta }_{ij}+b{m}_i{m}_j$, where ${\delta }_{ij}$ is the Kronecker delta. If we restrict attention to a change of shape with zero first-order volume change, its trace gives $3a+b=0$. Requiring the longitudinal component to equal ${\lambda }_s$ gives $a+b={\lambda }_s$. HenceThe tensor product $\mathbf{m}\otimes \mathbf{m}$ is symmetric and unchanged under $\mathbf{m}\mapsto -\mathbf{m}$. It uses two copies of the magnetic direction, so the third-rank isotropic tensor obstruction in isotropic linear magnetostriction vanishes does not apply. More generally, a small-magnetization expansion can contain $a{M}^2{\delta }_{ij}+b{M}_i{M}_j$; isotropy alone does not require its trace to vanish. The displayed saturated formula isolates the volume-preserving orientational contribution, rather than asserting that every material has zero volume magnetostriction.

Along a measuring unit vector $\mathbf{n}$, with $\mathbf{n}\cdot \mathbf{m}=\cos \theta$, the fractional length change isFor a stipulated ${\lambda }_s=20\times 10^{-6}$ and reference length $L=100\mathrm{mm}$, alignment gives $+2\mu \mathrm{m}$ and a perpendicular orientation gives $-1\mu \mathrm{m}$. Rotating from perpendicular to parallel therefore changes the length by $3\mu \mathrm{m}$, even though the saturated longitudinal value relative to the random reference is $2\mu \mathrm{m}$. A negative ${\lambda }_s$ reverses these elongation and contraction signs.

The eigenvalues of this infinitesimal strain tensor are ${\lambda }_s,-{\lambda }_s/2,-{\lambda }_s/2$, so $\Delta V/V=\mathrm{tr}{\epsilon }^{\mathrm{mag}}=0$ to first order. This is not exact finite incompressibility: using the illustrative deformation gradient $F=I+{\epsilon }^{\mathrm{mag}}$ givesA finite-deformation law would need higher-order corrections if exact volume conservation were required.

In a two-factor tensor product over a field, a pure tensor is also called a separable or decomposable tensor: it can be expressed as one product $v\otimes w$. Every tensor is a finite sum of pure tensors, but a sum need not itself be a pure tensor.^[ref]

For finite-dimensional vector spaces, the tensor-product coefficient matrix gives a decisive test: a nonzero tensor is a pure tensor exactly when its matrix rank is one. Indeed, every column of $v{w}^T$ is a scalar multiple of $v$. Conversely, if all columns of a nonzero matrix are multiples of one nonzero column $v$, collect their multipliers in $w$ to obtain $C=v{w}^T$. The zero tensor is also a pure tensor, for example $0\otimes w$, but its matrix rank is zero.

In ${\mathbb{R}}^2{\otimes }_{\mathbb{R}}{\mathbb{R}}^2$, the tensoris nonseparable: its matrix rank is two and its determinant is one, whereas $\det (v{w}^T)=0$. Directly, trying $t=(a{e}_1+b{e}_2)\otimes (c{f}_1+d{f}_2)$ requires $ac=bd=1$ and $ad=bc=0$. The first two equations force all four scalars nonzero, contradicting the others. In contrast, products sharing a factor do combine: $e_1\otimes f_1+e_1\otimes f_2=e_1\otimes (f_1+f_2)$. Thus being written as two summands alone does not establish nonseparability.

For finite free total chain complexes, evaluation identifies the underlying graded tensor product $X\otimes C^{\prime }$ with the graded Hom complex of chain complexes, where $X$ is the reversed dual chain complex. Under the precomposition-first Hom differential, conjugation by the sign $(-1{)}^{j(j-1)/2}$ makes this a chain isomorphism. The identity $\rho (j)-\rho (j-1)=j-1$ works for every integer degree. If the grading is unbounded and only degreewise finite, products on the Hom side need not equal direct sums on the tensor side.

For a map $f:E\to B$ with simply connected base and the usual finite-type hypotheses, choose a relative Sullivan algebra $M(B)\to M(B)\otimes \Lambda W$ modelling the path-space fibration, with acyclic total algebra. Tensor it over $M(B)$ with a model of $E$ via $f^{\ast }$. The resulting algebra models the homotopy fibre. Relative Sullivan extensions are cofibrant replacements, so this tensor product computes the derived algebraic pushout, contravariantly corresponding to the topological homotopy pullback. Linear differential pairs must be cancelled before reading rational homotopy groups from generators.

Take independent ensembles for two quantum channels and their Cartesian product ensemble. The Von Neumann entropy of a tensor product is the sum of the two entropies, both for its average output and for each labelled output. Thus its Holevo quantity is the sum of the two individual quantities. Taking ensembles arbitrarily close to each supremum proves the displayed inequality without requiring maximizing ensembles to exist. It does not assert additivity for arbitrary channels.

A super vector space is a vector space graded by the two-element group. Degree-preserving maps and the usual graded tensor product form a monoidal category. Over a field of characteristic different from two, this category has two different symmetries: ordinary interchange of tensor factors and interchange multiplied by $(-1{)}^{pq}$ on homogeneous factors of degrees $p,q$. The latter is the Koszul sign rule.

Objects are finite words of strand orientations. Morphisms are boundary-fixed ambient-isotopy classes of oriented tangles in a slab, composed by stacking compatible boundary words; horizontal juxtaposition is the tensor product. Cups, caps, identity strands and crossings generate the category. Ordinary unframed isotopy includes all three Reidemeister moves; framed variants impose different first-move relations.

A rank-$n$ tensor on a vector space $V$ is a multilinear map of $n$ vector or covector arguments, equivalently an element of an $n$-fold tensor product. Its components transform by one copy of the change-of-basis matrix for each index.

Realize $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$. Contract $r$ pairs of indices, one from each symmetric tensor, with the alternating invariant tensor $ϵ$, and symmetrize the remaining indices. This is a nonzero intertwining operator into $V_{m+n-2r}$. In the bihomogeneous polynomial realization, $(z_1{w}_2-z_2{w}_1{)}^r{z}_1^{m-r}{w}_1^{n-r}$ is a highest-weight vector of weight $m+n-2r$. Its lowering string has length $m+n-2r+1$. These inequivalent irreducible summands exhaust the tensor product, since their dimensions sum to $(m+1)(n+1)$.

A tensor field is a smooth section of a vector bundle formed from tensor products of the tangent bundle and cotangent bundle. With $k$ covector factors and $l$ vector factors it belongs toIts coefficients in any local frame vary smoothly. Different conventions list the two counts in different orders, so the factor order should be specified. Tensor contractions use the canonical evaluation of a covector on a vector and are independent of the frame.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

Tensor product need not commute with an infinite direct product. For example,The element $(1\mathrm{mod}n{)}_n$ has infinite order in the product, so it survives rational localization.

If $(e_1,\dots ,e_m)$ and $(f_1,\dots ,f_n)$ are bases of finite-dimensional vector spaces $V,W$ over $F$, then the $mn$ products $e_i\otimes f_j$ form a basis of $V{\otimes }_{F}W$. There is one independent coordinate for every pair of indices.^[ref]

For a proof outline, bilinearity expands each pure tensor asso these products form a spanning set. For linear independence, take the coordinate functionals $e^k,f^{\ell }$ from the two dual bases. The bilinear map $(v,w)\mapsto e^k(v)f^{\ell }(w)$ induces a linear functional ${\lambda }_{k\ell }$ on the tensor product, satisfying ${\lambda }_{k\ell }(e_i\otimes f_j)={\delta }_{ki}{\delta }_{\ell j}$. Applying it to a proposed zero linear combination selects just $c_{k\ell }$, forcing every coefficient to vanish. Thus dimensions multiply, rather than add. This finite-dimensional count includes a zero factor, whose tensor product has dimension zero.

An eigenstate with phase $\varphi$ becomes an eigenstate with phase $n\varphi$ under the tensor product of $n$ copies of the unitary operator. A promise $0\le \varphi <1/n$ prevents modulo-one aliasing. For $n=2^s$ and dyadic $\varphi =x/2^m$, exact quantum phase estimation then needs $m-s$ phase bits and $2^{m-s}-1$ queries to a supplied joint controlled oracle. This is a saving in joint-oracle calls, not in the total number of individual oracle invocations needed to build that joint operation.