14.20.1 Tensor Map Product Domain Verification

Ensuring compatibility in tensor map products through domain validation and structural alignment.

Tensor Map Product Domain Verification is the specific stage of the verification procedure concerned with confirming that a rule proposed on pairs $(v, w)$ actually descends to a well-defined function on the domain $V \otimes W$ , rather than merely on the product set $V \times W$ , by checking that the rule respects the bilinear relations used to construct the tensor product.

The Source of the Difficulty

A Rule on Pairs Is Not Automatically a Rule on Tensors

The tensor product $V \otimes W$ is built as a quotient of the free vector space on the set of pairs $V \times W$ by the subspace generated by relations such as

(v_{1} + v_{2}, w) - (v_{1}, w) - (v_{2}, w)

(α v, w) - α (v, w)

and the analogous relations in the second slot. A function assigned directly to pairs $(v, w)$ only descends to a function on the quotient $V \otimes W$ if it vanishes on this relation subspace, and domain verification is the check that this vanishing actually holds for the specific rule $(v, w) \mapsto f (v) \otimes g (w)$ .

Consequence of Skipping This Check

Without this verification, the same element of $V \otimes W$ , written as $v \otimes w$ in two different ways, for instance $(2 v) \otimes (\frac{1}{2} w)$ equal to $v \otimes w$ , could conceivably be sent to two different, inconsistent outputs by an unverified rule, which would make the rule not a function at all.

Checking Bilinearity of the Assignment

Additivity in the First Slot

Domain verification requires confirming

f (v_{1} + v_{2}) \otimes g (w) = f (v_{1}) \otimes g (w) + f (v_{2}) \otimes g (w)

which follows from linearity of $f$ , giving $f (v_{1} + v_{2}) = f (v_{1}) + f (v_{2})$ , together with bilinearity of the tensor symbol itself distributing this sum over the second factor.

Scalar Compatibility Across the Tensor Symbol

Domain verification also requires confirming

f (α v) \otimes g (w) = α (f (v) \otimes g (w)) = f (v) \otimes α g (w) = f (v) \otimes g (α w)

so that a scalar can be moved freely between the argument of $f$ , the coefficient position, and the argument of $g$ , again using linearity of $f$ and $g$ together with the defining scalar relation of the tensor symbol.

The Analogous Checks in the Second Slot

The same two checks, additivity and scalar compatibility, must be verified with the roles of $v$ and $w$ reversed, confirming that the rule respects the relations generated in the $W$ slot as well as those generated in the $V$ slot, since the tensor product's defining relations are imposed symmetrically on both factors.

Consequence: Existence of the Induced Map

The Universal Property Grants the Descent

Once all of these checks succeed, the universal property of the tensor product guarantees the existence of a unique linear map $V \otimes W \to V^{'} \otimes W^{'}$ agreeing with $(v, w) \mapsto f (v) \otimes g (w)$ on simple tensors, which is exactly $f \otimes g$ ; domain verification is therefore not a redundant formality but the actual content that makes the existence claim in the definition of $f \otimes g$ legitimate.

What Remains for Later Steps

Domain verification alone confirms that $f \otimes g$ is well defined as a function on all of $V \otimes W$ ; it does not by itself confirm that a separately proposed candidate map $h$ equals this function, which is the subject of the remaining steps of the tensor map product verification procedure.