Direct Sums, Tensor Products, and Index Structure

Direct sums and tensor products combine vector spaces in fundamentally different ways. A direct sum keeps independent sectors side by side; a tensor product linearizes simultaneous multilinear dependence. The distinction is structural, not notational: $V\oplus W$ is characterized by maps to and from its summands, whereas $V\otimes W$ is characterized by the fact that every bilinear map out of $V\times W$ factors uniquely through it. Bases turn these constructions into block arrays and multi-index arrays, but their universal properties explain which formulas are basis independent.

Required background. Vector Spaces, Duals, Linear Maps, and Bases supplies the linear maps, duals, bases, and change-of-basis rules used to state both universal properties.

This page concerns finite-dimensional vector spaces over one common field $\mathbb F=\mathbb R$ or $\mathbb C$. It uses algebraic tensor products. Infinite-dimensional Hilbert-space completions, operator domains, and topological tensor norms require additional choices and are not consequences of the finite-dimensional statements below.

Direct sums keep summands distinct

The direct sum of $V$ and $W$ is

$$V\oplus W = \{(v,w):v\in V,\ w\in W\},$$

with componentwise addition and scalar multiplication. It comes with canonical inclusions and projections,

$$\begin{aligned} \iota_V(v)&=(v,0), & \iota_W(w)&=(0,w),\\ \pi_V(v,w)&=v, & \pi_W(v,w)&=w. \end{aligned}$$

They obey

$$\pi_V\iota_V=\mathbb 1_V, \qquad \pi_W\iota_W=\mathbb 1_W, \qquad \iota_V\pi_V+\iota_W\pi_W=\mathbb 1_{V\oplus W},$$

while the crossed compositions vanish. Given maps $F_V:U\to V$ and $F_W:U\to W$, there is a unique map $F:U\to V\oplus W$ with $\pi_VF=F_V$ and $\pi_WF=F_W$, namely $F(u)=(F_V(u),F_W(u))$. Given maps $G_V:V\to U$ and $G_W:W\to U$, there is a unique map $G:V\oplus W\to U$ with $G\iota_V=G_V$ and $G\iota_W=G_W$, namely $G(v,w)=G_V(v)+G_W(w)$. These properties make the finite direct sum both a product and a coproduct in the category of vector spaces, although no category theory is needed to use them.

If $\{e_a\}$ and $\{f_i\}$ are bases of $V$ and $W$, then

$$\{(e_a,0)\}_a\cup\{(0,f_i)\}_i$$

is a basis of $V\oplus W$, so

$$\dim(V\oplus W)=\dim V+\dim W.$$

A linear map on a direct sum has a block form. For example,

$$T:V\oplus W\longrightarrow V\oplus W, \qquad T= \begin{pmatrix} T_{VV}&T_{VW}\\ T_{WV}&T_{WW} \end{pmatrix},$$

where the subscripts record actual source and target spaces: $T_{VW}:W\to V$, for instance. A block can be added to another block only when their types agree.

A direct sum is appropriate when alternatives or independently identified sectors are to remain visible. It does not create a slot for one vector from each summand. That simultaneous, bilinear role belongs to the tensor product.

The tensor product linearizes bilinear maps

The Cartesian product $V\times W$ is a set and a vector space under componentwise operations, but the map

$$(v,w)\longmapsto v\otimes w$$

is meant to be linear in each argument separately, not linear in the pair with the Cartesian-product vector-space structure. The tensor product $V\otimes W$ is a vector space equipped with a bilinear map

$$\otimes:V\times W\longrightarrow V\otimes W$$

having the following universal property: for every vector space $U$ and every bilinear map $B:V\times W\to U$, there is a unique linear map $\widetilde B:V\otimes W\to U$ such that

$$B(v,w)=\widetilde B(v\otimes w).$$

Equivalently,

$$\operatorname{Bil}(V,W;U) \cong \operatorname{Hom}(V\otimes W,U)$$

naturally in all three spaces. This is the defining feature of the tensor product. Any two constructions satisfying it are related by a unique isomorphism that preserves every elementary tensor $v\otimes w$. Conrad, n.d., §§ 2–5, PDF develops the quotient and universal-property constructions; Axler 2024, Chapter 3 supplies the surrounding finite-dimensional map and duality framework.

One explicit construction begins with the free vector space on formal pairs $(v,w)$ and quotients by the subspace generated by the bilinearity relations

$$\begin{aligned} (v+v',w)&-(v,w)-(v',w),\\ (v,w+w')&-(v,w)-(v,w'),\\ (\lambda v,w)&-\lambda(v,w),\\ (v,\lambda w)&-\lambda(v,w). \end{aligned}$$

The equivalence class of $(v,w)$ is written $v\otimes w$. The quotient immediately gives

$$\begin{aligned} (v+v')\otimes w&=v\otimes w+v'\otimes w,\\ v\otimes(w+w')&=v\otimes w+v\otimes w',\\ (\lambda v)\otimes w&=v\otimes(\lambda w) =\lambda(v\otimes w). \end{aligned}$$

An element of the form $v\otimes w$ is an elementary or pure tensor. Every tensor is a finite sum of pure tensors, but most tensors are not pure. Moreover, a decomposition into pure tensors is generally not unique.

Bases, dimensions, and decomposability

If $\{e_a\}_{a=1}^m$ is a basis of $V$ and $\{f_i\}_{i=1}^n$ is a basis of $W$, then

$$\{e_a\otimes f_i:1\leq a\leq m,\ 1\leq i\leq n\}$$

is a basis of $V\otimes W$. Consequently,

$$\dim(V\otimes W)=(\dim V)(\dim W),$$

and every tensor has a unique component expansion

$$T=T^{ai}e_a\otimes f_i.$$

For a pure tensor $v\otimes w$, the coefficient array factors:

$$T^{ai}=v^a w^i.$$

After bases are chosen, this is a rank-one matrix. Hence, for a nonzero two-factor tensor, purity is equivalent to matrix rank one. In particular, if $e_1,e_2$ and $f_1,f_2$ are independent, then

$$e_1\otimes f_1+e_2\otimes f_2$$

is not a pure tensor: its coefficient matrix has rank two. Although the matrix depends on the chosen bases, its rank does not, because a basis change multiplies it on the left and right by invertible matrices.

This rank criterion is special to a bipartite tensor viewed as a matrix. For three or more tensor factors, tensor rank has subtler behavior and is not captured by a single matrix rank.

Tensoring maps and canonical rearrangements

Given linear maps $A:V\to V'$ and $B:W\to W'$, bilinearity of $(v,w)\mapsto A(v)\otimes B(w)$ produces a unique linear map

$$A\otimes B:V\otimes W\longrightarrow V'\otimes W'$$

satisfying

$$(A\otimes B)(v\otimes w)=A(v)\otimes B(w).$$

It respects identities and composition:

$$\mathbb 1_V\otimes\mathbb 1_W = \mathbb 1_{V\otimes W},$$ $$(A'\otimes B')(A\otimes B) = (A'A)\otimes(B'B),$$

whenever the compositions are defined. These rules follow from agreement on all pure tensors, which span the tensor product.

There are also canonical isomorphisms

$$\begin{aligned} (U\otimes V)\otimes W &\longrightarrow U\otimes(V\otimes W), & (u\otimes v)\otimes w &\longmapsto u\otimes(v\otimes w),\\ V\otimes W &\longrightarrow W\otimes V, & v\otimes w &\longmapsto w\otimes v,\\ \mathbb F\otimes V &\longrightarrow V, & \lambda\otimes v &\longmapsto \lambda v. \end{aligned}$$

They justify suppressing parentheses in a multiple tensor product once the order of factors is clear. “Canonical” does not mean literal equality: the spaces are linked by specified natural isomorphisms.

Tensor products distribute over direct sums:

$$(V\oplus W)\otimes U \cong (V\otimes U)\oplus(W\otimes U),$$

with $(v,w)\otimes u\mapsto(v\otimes u,w\otimes u)$. Thus the two constructions cooperate without becoming interchangeable.

Multilinear forms, maps, and index types

The universal property identifies a bilinear form $B:V\times W\to\mathbb F$ with a linear functional on $V\otimes W$:

$$(V\otimes W)^* \cong \operatorname{Bil}(V,W;\mathbb F).$$

Because the spaces here are finite dimensional, there is also a canonical isomorphism

$$V^*\otimes W^* \longrightarrow (V\otimes W)^*, \qquad \alpha\otimes\beta \longmapsto \bigl[v\otimes w\mapsto\alpha(v)\beta(w)\bigr].$$

The finite-dimensional hypothesis matters: for topological vector spaces, both the choice of tensor completion and the choice of continuous dual enter.

Similarly,

$$W\otimes V^* \cong \operatorname{Hom}(V,W), \qquad w\otimes\alpha \longmapsto \bigl[v\mapsto\alpha(v)w\bigr].$$

In particular, $V\otimes V^*\cong\operatorname{End}(V)$. The tensor

$$\mathbb 1_V=e_a\otimes e^a$$

corresponds to the identity map. Here the repeated index denotes summation over the dual basis; evaluation occurs when the factor $e^a$ acts on the input vector under the canonical map to $\operatorname{End}(V)$.

More generally, a tensor of type $(r,s)$ on $V$ is an element of

$$V^{\otimes r}\otimes(V^*)^{\otimes s}.$$

Its upper indices label vector factors and its lower indices label dual factors. If

$$e'_a=e_bM^b{}_a, \qquad f'_i=f_jN^j{}_i,$$

then the components of $T\in V\otimes W$ transform as

$$T'^{ai} = (M^{-1})^a{}_b (N^{-1})^i{}_j T^{bj}.$$

Each factor contributes its own transformation. A contraction is basis independent only when it evaluates a vector factor against a matching dual factor. For $A=A^a{}_b e_a\otimes e^b\in V\otimes V^*$,

$$\operatorname{tr}A=A^a{}_a$$

is such a contraction. Turning two upper indices into an upper and a lower one requires a specified nondegenerate form; index position alone does not supply that map. Frankel 2012, §§ 2.1–2.4 develops this tensor-type and contraction discipline in coordinate and invariant language.

Symmetric and antisymmetric parts

When the two factors are the same space, the swap map

$$\tau:V\otimes V\longrightarrow V\otimes V, \qquad \tau(v\otimes w)=w\otimes v$$

satisfies $\tau^2=\mathbb 1$. Over $\mathbb R$ or $\mathbb C$, the operators

$$P_{\mathrm{sym}} = \frac12(\mathbb 1+\tau), \qquad P_{\mathrm{alt}} = \frac12(\mathbb 1-\tau)$$

are complementary projectors. Therefore

$$V\otimes V = \operatorname{Sym}^2V \oplus \Lambda^2V,$$

where the two summands are the $+1$ and $-1$ eigenspaces of $\tau$. If $\dim V=n$, then

$$\dim\operatorname{Sym}^2V=\frac{n(n+1)}2, \qquad \dim\Lambda^2V=\frac{n(n-1)}2.$$

This alternating subspace is canonically identified with the quotient-defined exterior square used on the graded-algebra page by

$$v\wedge w \longmapsto P_{\mathrm{alt}}(v\otimes w) = \frac12(v\otimes w-w\otimes v).$$

For more factors, permutations act by rearranging tensor slots. Symmetric, antisymmetric, and more general index symmetries are obtained by projecting onto appropriate permutation types. The exterior-algebra and sign conventions needed for fermionic variables are developed later on Exterior and Graded Algebra, Grassmann Variables, and Berezin Integration.

QFT-facing example: multiplets and multiparticle sectors

The tensor-product construction of multiparticle state spaces used here is the finite-dimensional algebraic model of the Fock-space discussion in Schwartz 2014, Chapter 2.

Let a group $G$ act on finite-dimensional internal spaces $R$ and $S$ by representations $\rho_R$ and $\rho_S$. A pair of degrees of freedom transforms on the tensor product through

$$\rho_{R\otimes S}(g) = \rho_R(g)\otimes\rho_S(g).$$

A bilinear contraction $C:R\times S\to\mathbb F$ is equivalently $\widetilde C\in(R\otimes S)^*$. It is invariant when

$$\widetilde C\, \bigl(\rho_R(g)\otimes\rho_S(g)\bigr) = \widetilde C$$

for every $g\in G$. In components, a coupling $C_{ai}\phi^a\chi^i$ is therefore a scalar only after the spaces and the transformation law of $C$ have been specified. An arbitrary repeated-index pattern does not guarantee invariance.

For a finite-dimensional one-particle approximation $\mathcal H_1$, the ordered $n$-slot space is

$$\mathcal H_1^{\otimes n}.$$

Identical bosonic and fermionic kinematics select the symmetric and antisymmetric subspaces,

$$\operatorname{Sym}^n\mathcal H_1, \qquad \Lambda^n\mathcal H_1,$$

and the corresponding finite-particle organization has the algebraic form

$$\begin{aligned} \mathcal F_+^{\mathrm{alg}}(\mathcal H_1) &= \bigoplus_{n=0}^{\infty}\operatorname{Sym}^n\mathcal H_1,\\ \mathcal F_-^{\mathrm{alg}}(\mathcal H_1) &= \bigoplus_{n=0}^{\infty}\Lambda^n\mathcal H_1, \end{aligned}$$

with $\operatorname{Sym}^0\mathcal H_1=\Lambda^0\mathcal H_1=\mathbb F$.

This formula displays both constructions: tensor powers describe simultaneous particle slots, while the direct sum keeps different particle-number sectors distinct. It does not by itself establish the spin–statistics connection, choose normalizations, or construct the completed Fock Hilbert space. Those physical and analytic steps belong to Multiparticle States, Statistics, and Fock Organization.

Common pitfalls

Confusing $V\otimes W$ with $V\times W$. The Cartesian product packages a pair and is isomorphic to the finite direct sum. The tensor product is generated by bilinear symbols and has product, rather than sum, dimension.

Assuming every tensor is pure. Pure tensors span $V\otimes W$, but a sum of pure tensors usually cannot be collapsed to one pure tensor. In two factors, coefficient-matrix rank detects the obstruction.

Treating canonical isomorphisms as literal identities. Associativity, symmetry, and distributivity maps are natural and unambiguous, but they still rearrange specified factors. Suppressing them is safe only after the factor order and types have been fixed.

Importing finite-dimensional dual formulas unchanged. The isomorphism $V^*\otimes W^*\cong(V\otimes W)^*$ is automatic in the present finite-dimensional setting. Infinite-dimensional continuous duals and completed tensor products require extra hypotheses and may not obey the same formula.

Symmetrizing distinguishable slots without a reason. The tensor product itself records ordered factors. Symmetry or antisymmetry is an additional projection justified by the physical or mathematical problem.

Exercises

1. Let $V=W=\mathbb R^2$ with bases $e_1,e_2$ and $f_1,f_2$. Decide which of the following tensors are pure:

   $$T_1=e_1\otimes f_1+2e_1\otimes f_2, \qquad T_2=e_1\otimes f_1+e_2\otimes f_2.$$
   Solution

   The first tensor factors:

   $$T_1=e_1\otimes(f_1+2f_2),$$

   so it is pure. The coefficient matrix of $T_2$ is the $2\times2$ identity, which has rank two. Therefore $T_2$ is not pure.

2. Show that the rule $B(v,w)=\alpha(v)\beta(w)u$ defines a bilinear map $V\times W\to U$, and write its linear factorization through $V\otimes W$.

   Solution

   Linearity of $\alpha$ and $\beta$ gives linearity in each argument. Universality produces

   $$\widetilde B:V\otimes W\longrightarrow U, \qquad \widetilde B(v\otimes w) =\alpha(v)\beta(w)u.$$

   On a general tensor $\sum_kv_k\otimes w_k$, linearity gives $\widetilde B(\sum_kv_k\otimes w_k) =\sum_k\alpha(v_k)\beta(w_k)u$.

3. For $\dim V=n$, count bases for the symmetric and antisymmetric two-tensors and recover their dimensions.

   Solution

   A basis for $\operatorname{Sym}^2V$ consists of $e_a\otimes e_a$ and $e_a\otimes e_b+e_b\otimes e_a$ with $a<b$. There are $n+n(n-1)/2=n(n+1)/2$ elements. A basis for $\Lambda^2V$ consists of $e_a\otimes e_b-e_b\otimes e_a$ with $a<b$, giving $n(n-1)/2$ elements.

4. If $A:V\to V'$ and $B:W\to W'$ are invertible, show that $A\otimes B$ is invertible and identify its inverse.

   Solution

   Functoriality gives

   $$(A^{-1}\otimes B^{-1})(A\otimes B) = (A^{-1}A)\otimes(B^{-1}B) = \mathbb 1_{V\otimes W}.$$

   The reverse composition is the identity on $V'\otimes W'$, so $(A\otimes B)^{-1}=A^{-1}\otimes B^{-1}$.

References

• Sheldon Axler, Linear Algebra Done Right, 4th ed., Springer, 2024, Chapters 1 and 3, for direct sums, linear maps, duality, and the finite-dimensional framework.
• Keith Conrad, Tensor Products, PDF, University of Connecticut lecture notes, n.d., §§ 2–5, for the quotient construction, universal property, bases, and non-elementary tensors.
• Theodore Frankel, The Geometry of Physics, 3rd ed., Cambridge University Press, 2012, §§ 2.1–2.4, for tensor types, contractions, and coordinate transformations.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, Chapter 2, for one-particle and multiparticle state spaces in relativistic field theory.