Questions: Tensor Products as Universal Constructions

Question 1

Let V be a 3-dimensional vector space and W a 4-dimensional vector space. A student claims every element of V ⊗ W can be written as v ⊗ w for some v ∈ V and w ∈ W. What is wrong with this claim?

• A. Nothing — pure tensors span V ⊗ W, so every element is a pure tensor
• B. General elements of V ⊗ W are linear combinations of pure tensors; most such combinations are not pure tensors themselves
• C. The claim fails because dim(V ⊗ W) = 12 exceeds dim(V) + dim(W) = 7, making pure tensors insufficient as a basis
• D. Pure tensors v ⊗ w don't exist when V and W have different dimensions

Answer: (B)

Pure tensors v ⊗ w *do* span V ⊗ W (every element is a sum of pure tensors), but 'span' does not mean 'equal.' A general element looks like Σcᵢⱼ(eᵢ ⊗ fⱼ) — a linear combination of basis pure tensors — and this combination is typically not itself a pure tensor. For instance, e₁ ⊗ f₁ + e₂ ⊗ f₂ in ℝ² ⊗ ℝ² cannot be written as a single a ⊗ b for any a ∈ ℝ², b ∈ ℝ². This 'entangled' element is the categorical analog of quantum entanglement, which is precisely the same phenomenon.

Question 2

The universal property of the tensor product states that Hom(A ⊗ B, C) ≅ Bilin(A × B, C). What problem does this isomorphism solve?

• A. It shows that bilinear maps factor through the cartesian product A × B, confirming A × B is the right universal object
• B. It converts the problem of specifying a bilinear map A × B → C into the equivalent problem of specifying an ordinary linear map A ⊗ B → C, linearizing the bilinearity
• C. It classifies all maps from A to B by composing with elements of C
• D. It proves that tensor products and direct products are isomorphic for finite-dimensional vector spaces

Answer: (B)

The isomorphism is a 'linearization' device: bilinear maps are harder to work with categorically because they are not morphisms in the usual sense. The tensor product converts the problem — instead of tracking a bilinear map out of A × B, you track a linear map out of A ⊗ B. Since linear maps are the morphisms of the category, all the machinery of category theory becomes available. This is the same move as converting a multilinear problem into a linear one in linear algebra: you encode the complexity into the object (A ⊗ B) rather than the map.

Question 3

The cartesian product A × B already serves as the universal object for bilinear maps from A × B to C.

• T. True
• F. False

Answer: False

This is the precise misconception the tensor product is designed to correct. The cartesian product A × B is the universal object for *pairs* of linear maps — given f: X → A and g: X → B, there is a unique linear map into A × B. But a bilinear map f: A × B → C is linear in each variable separately, not linear on the product as a whole. These are different conditions: f(a + a', b) ≠ f(a, b) + f(a', b) in general for a map that is merely linear on A × B. The cartesian product does not represent bilinear maps, which is why the tensor product must be introduced as a new construction.

Question 4

The tensor product A ⊗ B is characterized uniquely up to unique isomorphism by its universal property.

• T. True
• F. False

Answer: True

This is the standard consequence of universal properties in category theory. Any two objects satisfying the same universal property are related by a unique isomorphism — they are 'the same' in the categorical sense. For tensor products, if T and T' both come with canonical bilinear maps with the same universal property, then the universal property of T produces a unique linear map T → T', and vice versa, and these compose to the identity. This is the reason universal properties are used to *define* constructions like tensor products: the definition specifies behavior, not implementation, and any implementation satisfying the behavior is equivalent.

Question 5

Why can't the cartesian product A × B serve as the universal object for bilinear maps, and what does the tensor product A ⊗ B do differently to solve this problem?

Model answer: A bilinear map f: A × B → C is linear in each variable separately — f(a + a', b) = f(a,b) + f(a',b) and f(λa, b) = λf(a,b) — but this is not the same as f being linear on A × B as a whole. The cartesian product represents pairs of independent maps into A and B, not this mixed bilinearity. The tensor product A ⊗ B is defined specifically so that bilinear maps out of A × B correspond bijectively to linear maps out of A ⊗ B: any bilinear f: A × B → C factors uniquely as f = f̃ ∘ ⊗, where ⊗: A × B → A ⊗ B is the canonical bilinear map and f̃: A ⊗ B → C is linear. The tensor product encodes the bilinearity into its structure so that what remains — maps out of A ⊗ B — are ordinary linear maps.

The key phrase is 'linearizes bilinearity.' The tensor product's universal property says: to give a bilinear map out of A × B is the same as giving a linear map out of A ⊗ B. The 'hard' structure (bilinearity) is absorbed into the definition of A ⊗ B, leaving only the 'easy' structure (linearity) exposed. This is a general categorical strategy: represent complex map types by constructing objects that make those maps into ordinary morphisms.