Physlib.Relativity.Tensors.Constructors

The $k$-linear equivalence between the object $S.FD.obj \{as := c\}$ and the space of pure tensors $\text{Pure } S ![c]$ for a single index $c \in C$.

The $k$-linear equivalence between the object $S.FD.obj \{as := c\}$ and the tensor space $S.Tensor ![c]$ for a single index $c \in C$.

For any index $c \in C$ and any pure tensor $p \in \text{Pure } S ![c]$ of rank 1, the following identity holds: $$\text{fromSingleT}^{-1}(p.\text{toTensor}) = \text{fromSingleP}^{-1}(p)$$ where $\text{fromSingleT}^{-1}$ is the inverse of the $k$-linear equivalence between the vector space $S.FD.obj \{ as := c \}$ and the tensor space $S.Tensor ![c]$, and $\text{fromSingleP}^{-1}$ is the inverse of the $k$-linear equivalence between the same vector space and the space of pure tensors $\text{Pure } S ![c]$.

For any index $c \in C$ and any vector $x \in S.FD.obj \{ as := c \}$, the tensor constructed from $x$ via the $k$-linear equivalence $\text{fromSingleT}$ is equal to the rank-1 pure tensor associated with $x$. Specifically, $$\text{fromSingleT}(x) = \text{Pure.toTensor}(\iota_x)$$ where $\iota_x \in \text{Pure } S ![c]$ is the pure tensor mapping the unique index $0$ to the vector $x$.

Let $S$ be a tensor species over a group $G$ and an index set $C$. For any index $c \in C$, let $V_c$ be the vector space $S.FD.obj \{ as := c \}$ and let $\rho$ be the representation of $G$ on $V_c$. For any vector $x \in V_c$ and any group element $g \in G$, the action of $g$ on the rank-1 tensor constructed from $x$ is equal to the tensor constructed from the transformed vector $\rho(g)x$. Specifically, $$g \cdot \text{fromSingleT}(x) = \text{fromSingleT}(\rho(g)x)$$ where $\text{fromSingleT}$ is the linear equivalence that maps a vector to its corresponding rank-1 tensor.

For any indices $c, c_1 \in C$, let $h: \{c\} = \{c_1\}$ be an equality in the discrete category of indices. For any vector $x \in V_c$ (where $V_c$ denotes the object $S.FD.obj \{as := c\}$), the following equality holds: $$\text{fromSingleT}(S.FD.map(h)(x)) = \text{permT}_{\text{id}}(\text{fromSingleT}(x))$$ where $S.FD.map(h): V_c \to V_{c_1}$ is the linear map induced by the equality $h$, and $\text{permT}_{\text{id}}$ is the tensor permutation map corresponding to the identity permutation, which re-indexes the tensor from $c$ to $c_1$.

Let $S$ be a tensor species over a ring $k$ and $C$ be the set of index colors. For any index $c \in C$, let $V_c$ denote the $k$-module associated with $c$ (formally $S.FD.obj \{as := c\}$) and let $\tau(c)$ be the index color such that a contraction exists between $V_c$ and $V_{\tau(c)}$. Given vectors $x \in V_c$ and $y \in V_{\tau(c)}$, the contraction of the tensor product of the two rank-1 tensors constructed from $x$ and $y$ is equal to the scalar pairing of $x$ and $y$ (represented as a rank-0 tensor). Specifically, $$\text{contr}_{0,0} (\text{fromSingleT}(x) \otimes \text{fromSingleT}(y)) = \langle x, y \rangle_c \cdot \mathbf{1}$$ where: - $\text{fromSingleT}$ maps a vector to its corresponding rank-1 tensor. - $\otimes$ denotes the tensor product of tensors (`prodT`). - $\text{contr}_{0,0}$ is the contraction operation `contrT` acting on the indices of the rank-2 tensor. - $\langle \cdot, \cdot \rangle_c$ is the contraction pairing defined by the tensor species $S$ for color $c$. - $\mathbf{1}$ is the identity rank-0 tensor (`Pure.toTensor default`).

Given a tensor species $S$ over a ring $k$ and index colors $c_1, c_2 \in C$, let $V_{c_1}$ and $V_{c_2}$ be the $k$-modules (vector spaces) associated with these colors. The function `fromPairT` is a $k$-linear map from the tensor product $V_{c_1} \otimes_k V_{c_2}$ to the space of rank-2 tensors $S.\text{Tensor}(c_1, c_2)$. It constructs a two-index tensor by taking the tensor product of the single-index tensors corresponding to the elements in $V_{c_1}$ and $V_{c_2}$.

For a tensor species $S$ over a ring $k$ and index colors $c_1, c_2 \in C$, let $x \in V_{c_1}$ and $y \in V_{c_2}$ be elements of the vector spaces associated with these colors. The rank-2 tensor constructed from their tensor product $x \otimes_k y$ via the map `fromPairT` is equal to the tensor product `prodT` of the rank-1 tensors `fromSingleT x` and `fromSingleT y`, subject to an identity permutation of indices to reconcile the index type definitions.

Let $S$ be a tensor species over a group $G$ and a ring $k$. For any index colors $c_1, c_2 \in C$, let $V_{c_1}$ and $V_{c_2}$ be the $k$-modules associated with these colors, and let $\rho$ denote the representation of $G$ on these modules. For any element $x \in V_{c_1} \otimes_k V_{c_2}$ and any group element $g \in G$, the action of $g$ on the rank-2 tensor constructed from $x$ satisfies: $$g \cdot \text{fromPairT}(x) = \text{fromPairT}((\rho(g) \otimes \rho(g))(x))$$ where $\text{fromPairT}$ is the $k$-linear map that constructs a rank-2 tensor from the tensor product of two vector spaces, and $(\rho(g) \otimes \rho(g))$ is the induced action on the tensor product $V_{c_1} \otimes_k V_{c_2}$.

For a tensor species $S$ over a ring $k$ and index colors $c_1, c_2, c_2' \in C$, let $V_{c}$ denote the $k$-module (vector space) associated with color $c$. Let $h: c_2 = c_2'$ be an equality and let $f: V_{c_2} \to V_{c_2'}$ be the $k$-linear map induced by $h$ (the map $S.FD.map(h)$). For any element $x \in V_{c_1} \otimes_k V_{c_2}$, the rank-2 tensor constructed from the mapped element satisfies: $$\text{fromPairT}((\text{id} \otimes f)x) = \text{permT}_{\text{id}}(\text{fromPairT } x)$$ where $\text{fromPairT}$ is the $k$-linear map that constructs a rank-2 tensor from a tensor product of vector spaces, and $\text{permT}_{\text{id}}$ is the tensor permutation map corresponding to the identity permutation, which re-indexes the tensor to account for the color change from configuration $[c_1, c_2]$ to $[c_1, c_2']$.

For a tensor species $S$ over a ring $k$ and index colors $c_1, c_2 \in C$, let $V_{c_1}$ and $V_{c_2}$ be the $k$-modules (vector spaces) associated with these colors. Given an element $x \in V_{c_1} \otimes_k V_{c_2}$, let $\text{comm}: V_{c_1} \otimes_k V_{c_2} \cong V_{c_2} \otimes_k V_{c_1}$ be the canonical commutativity isomorphism of the tensor product. The theorem states that applying the rank-2 tensor construction map `fromPairT` to the commuted element $\text{comm}(x)$ is equivalent to taking the tensor `fromPairT(x)` and permuting its indices by the swap $\sigma(0)=1, \sigma(1)=0$. That is: $$\text{fromPairT}(\text{comm}(x)) = \text{perm}_T(\sigma, h)(\text{fromPairT}(x))$$ where $\sigma = [1, 0]$ is the permutation swapping the two indices and $h$ is the condition ensuring the consistency of the index colors.

Given a vector $x$ in the space $V_c$ and a tensor $y$ in the tensor product $V_{\tau(c)} \otimes_k V_{c_2}$, where $\tau(c)$ denotes the dual index of $c$, this function constructs a rank-1 tensor in $S.\text{Tensor}([c_2])$ by contracting $x$ with the first index of $y$. Categorically, the process is defined by: 1. Forming the tensor product $x \otimes y \in V_c \otimes_k (V_{\tau(c)} \otimes_k V_{c_2})$. 2. Applying the inverse associator $\alpha^{-1}$ to regroup the terms as $(V_c \otimes_k V_{\tau(c)}) \otimes_k V_{c_2}$. 3. Applying the contraction morphism $\text{contr}_c : V_c \otimes_k V_{\tau(c)} \to k$ to the first two components. 4. Using the left unitor $\lambda : k \otimes_k V_{c_2} \cong V_{c_2}$ to obtain a vector in $V_{c_2}$. 5. Converting this resulting vector into a single-indexed tensor via the linear equivalence `fromSingleT`. For simple tensors $y = y_1 \otimes y_2$, the operation satisfies: $$\text{fromSingleTContrFromPairT}(x, y_1 \otimes y_2) = \text{contr}_c(x \otimes y_1) \cdot \text{fromSingleT}(y_2)$$ where $\cdot$ denotes scalar multiplication by the result of the contraction.

For any indices $c, c_2 \in C$, let $V_c$ and $V_{c_2}$ be the corresponding vector spaces and $V_{\tau(c)}$ be the vector space associated with the dual index of $c$. Given a vector $x \in V_c$, and vectors $y_1 \in V_{\tau(c)}$ and $y_2 \in V_{c_2}$ forming the simple tensor $y_1 \otimes y_2$, the contraction operation $\text{fromSingleTContrFromPairT}$ satisfies: $$\text{fromSingleTContrFromPairT}(x, y_1 \otimes y_2) = \text{contr}_c(x \otimes y_1) \cdot \text{fromSingleT}(y_2)$$ where $\text{contr}_c : V_c \otimes V_{\tau(c)} \to k$ is the contraction morphism, $\text{fromSingleT}$ is the linear equivalence mapping a vector to a rank-1 tensor, and $\cdot$ denotes scalar multiplication.

Let $S$ be a tensor species over a ring $k$ and $C$ be the set of index colors. For any index colors $c, c_2 \in C$, let $V_c$ be the $k$-module associated with $c$ and $V_{\tau c}$ be the $k$-module associated with the dual color of $c$. Given a vector $x \in V_c$ and vectors $y_1 \in V_{\tau c}, y_2 \in V_{c_2}$, the contraction of the rank-3 tensor product of $\text{fromSingleT}(x)$ and $\text{fromPairT}(y_1 \otimes y_2)$ is equal to the rank-1 tensor $\text{fromSingleTContrFromPairT}(x, y_1 \otimes y_2)$. Specifically, $$\text{contr}_{0,1} (\text{fromSingleT}(x) \otimes \text{fromPairT}(y_1 \otimes y_2)) = \text{permT}_{\text{id}} (\text{fromSingleTContrFromPairT}(x, y_1 \otimes y_2))$$ where: - $\text{fromSingleT}$ and $\text{fromPairT}$ are the canonical maps from vectors/tensor products of vectors to rank-1 and rank-2 tensors, respectively. - $\text{prodT}$ (denoted by $\otimes$) forms the rank-3 tensor with indices $[c, \tau c, c_2]$. - $\text{contr}_{0,1}$ is the contraction operation `contrT` acting on the indices of color $c$ and $\tau c$ (the indices at positions 0 and 1). - $\text{fromSingleTContrFromPairT}(x, y_1 \otimes y_2)$ represents the vector contraction $\langle x, y_1 \rangle_c$ resulting in the rank-1 tensor $\langle x, y_1 \rangle_c \cdot \text{fromSingleT}(y_2)$. - $\text{permT}_{\text{id}}$ is the identity permutation used for index type alignment.

Let $S$ be a tensor species over a ring $k$ and $C$ be the set of index colors. For any index colors $c, c_2 \in C$, let $V_c$ be the $k$-module associated with $c$ and $V_{\tau c}$ be the $k$-module associated with the dual color of $c$. Given a vector $x \in V_c$ and a tensor $y \in V_{\tau c} \otimes_k V_{c_2}$, the contraction of the rank-3 tensor formed by the product of the rank-1 tensor $\text{fromSingleT}(x)$ and the rank-2 tensor $\text{fromPairT}(y)$ is equal to the rank-1 tensor $\text{fromSingleTContrFromPairT}(x, y)$. Specifically, $$\text{contr}_{0,1} (\text{fromSingleT}(x) \otimes \text{fromPairT}(y)) = \text{permT}_{\text{id}} (\text{fromSingleTContrFromPairT}(x, y))$$ where: - $\text{fromSingleT}$ and $\text{fromPairT}$ are the canonical maps from vectors and tensor products of vectors to rank-1 and rank-2 tensors, respectively. - $\text{prodT}$ (denoted by $\otimes$) forms the rank-3 tensor with indices $[c, \tau c, c_2]$. - $\text{contr}_{0,1}$ is the contraction operation `contrT` acting on the indices at positions 0 and 1 (colors $c$ and $\tau c$). - $\text{fromSingleTContrFromPairT}(x, y)$ is the rank-1 tensor in $S.\text{Tensor}([c_2])$ resulting from the contraction of $x$ with the first index of $y$. - $\text{permT}_{\text{id}}$ is the identity permutation used for index type alignment.

Given a tensor species $S$ over a ring $k$ and index colors $c, c_1, c_2 \in C$, let $V_{c_i}$ be the $k$-modules associated with these colors. For elements $x \in V_{c_1} \otimes_k V_c$ and $y \in V_{\tau c} \otimes_k V_{c_2}$ (where $\tau c$ is the dual color of $c$), this function constructs a rank-2 tensor in $S.\text{Tensor}(c_1, c_2)$. The construction is performed categorically: it takes the tensor product $x \otimes y$, reassociates the factors to isolate $V_c \otimes_k V_{\tau c}$, applies the contraction morphism $V_c \otimes_k V_{\tau c} \to k$, and uses the left unitor to obtain an element in $V_{c_1} \otimes_k V_{c_2}$, which is finally mapped to a rank-2 tensor via `fromPairT`.

Let $S$ be a tensor species over a ring $k$, and for any color $c \in C$, let $V_c$ be the $k$-module associated with it. For any vectors $x_1 \in V_{c_1}$, $x_2 \in V_c$, $y_1 \in V_{\tau c}$, and $y_2 \in V_{c_2}$ (where $\tau c$ is the dual color of $c$), the rank-2 tensor obtained by the contraction operation `fromPairTContr` on the elementary tensors $x_1 \otimes x_2$ and $y_1 \otimes y_2$ satisfies: $$\text{fromPairTContr}(x_1 \otimes x_2, y_1 \otimes y_2) = \text{contr}_c(x_2 \otimes y_1) \cdot \text{fromPairT}(x_1 \otimes y_2)$$ where $\text{contr}_c: V_c \otimes_k V_{\tau c} \to k$ is the contraction morphism for the color $c$, and $\text{fromPairT}$ maps an element of $V_{c_1} \otimes_k V_{c_2}$ to a rank-2 tensor.

Let $S$ be a tensor species over a ring $k$. For index colors $c, c_1, c_2 \in C$, let $V_{c_1}, V_c, V_{\tau c}$, and $V_{c_2}$ be the $k$-modules associated with these colors, where $\tau c$ is the dual color of $c$. For any vectors $x_1 \in V_{c_1}, x_2 \in V_c, y_1 \in V_{\tau c}$, and $y_2 \in V_{c_2}$, the contraction of the tensor product of the rank-2 tensors constructed from the elementary tensors $x_1 \otimes_k x_2$ and $y_1 \otimes_k y_2$ at the middle indices is equal to the rank-2 tensor obtained via the direct contraction map `fromPairTContr`. Specifically, $$\text{contr}_{1,2} \left( \text{fromPairT}(x_1 \otimes_k x_2) \otimes \text{fromPairT}(y_1 \otimes_k y_2) \right) = \text{fromPairTContr}(x_1 \otimes_k x_2, y_1 \otimes_k y_2)$$ where: - $\text{fromPairT}$ is the linear map from $V_a \otimes_k V_b$ to the space of rank-2 tensors. - $\otimes$ between tensors denotes the tensor product operation `prodT`. - $\text{contr}_{1,2}$ (formally `contrT 2 1 2`) denotes the contraction of the second index of the first tensor with the first index of the second tensor (indices 1 and 2 of the resulting rank-4 product). - The equality holds up to a canonical re-indexing of the indices (identity permutation).

Let $S$ be a tensor species over a ring $k$. For any index colors $c, c_1, c_2 \in C$, let $V_{c_1}, V_c, V_{\tau c}$, and $V_{c_2}$ be the $k$-modules associated with these colors, where $\tau c$ denotes the dual color of $c$. For any elements $x \in V_{c_1} \otimes_k V_c$ and $y \in V_{\tau c} \otimes_k V_{c_2}$, the contraction of the tensor product of the rank-2 tensors constructed from $x$ and $y$ is equal to the rank-2 tensor obtained via the direct contraction map `fromPairTContr`. Specifically, $$\text{contr}_{1,2} \left( \text{fromPairT}(x) \otimes \text{fromPairT}(y) \right) = \text{fromPairTContr}(x, y)$$ where: - $\text{fromPairT}$ is the $k$-linear map from $V_a \otimes_k V_b$ to the space of rank-2 tensors. - $\otimes$ between tensors denotes the tensor product operation `prodT`. - $\text{contr}_{1,2}$ (formally `contrT 2 1 2`) denotes the contraction of the second index of the first tensor with the first index of the second tensor (corresponding to the matching colors $c$ and $\tau c$). - The equality holds up to a canonical re-indexing (permutation) of the resulting indices.

Let $S$ be a tensor species over a ring $k$, and let $c, c_1 \in C$ be index colors with corresponding $k$-modules $V_c$ and $V_{c_1}$. For any element $x \in V_c \otimes_k V_{c_1}$ and any multi-index $b = (b_0, b_1)$ for rank-2 tensors of colors $(c, c_1)$, the coefficient of the rank-2 tensor $\text{fromPairT}(x)$ at index $b$ with respect to the tensor basis is equal to the coefficient of $x$ at the index pair $(b_0, b_1)$ with respect to the induced tensor product basis of $V_c \otimes_k V_{c_1}$. That is, $$[\text{fromPairT}(x)]_b = [x]_{(b_0, b_1)}.$$

Let $S$ be a tensor species over a ring $k$. For any index colors $c, c_1 \in C$ and basis indices $b_0 \in \{0, \dots, \text{dim}(V_c)-1\}$ and $b_1 \in \{0, \dots, \text{dim}(V_{c_1})-1\}$, let $e_{b_0}^{(c)}$ and $e_{b_1}^{(c_1)}$ be the basis vectors of the $k$-modules $V_c$ and $V_{c_1}$ respectively. Then, the value of the $k$-linear map $\text{fromPairT}$ on the tensor product of these basis vectors is the basis element of the rank-2 tensor space $S.\text{Tensor}(c, c_1)$ corresponding to the multi-index $(b_0, b_1)$: $$\text{fromPairT}(e_{b_0}^{(c)} \otimes_k e_{b_1}^{(c_1)}) = \mathcal{E}_{(b_0, b_1)},$$ where $\mathcal{E}_{(b_0, b_1)}$ denotes the basis element of the tensor space for the color pair $(c, c_1)$ at indices $(b_0, b_1)$.

Let $S$ be a tensor species over a ring $k$ and a group $G$. For any two colors $c_1, c_2 \in C$, let $V_{c_1}$ and $V_{c_2}$ be the $k$-modules (representations) associated with these colors. The function `fromConstPair` constructs a rank-2 tensor in $S.\text{Tensor}(c_1, c_2)$ from a morphism $v: \mathbb{1} \to V_{c_1} \otimes_k V_{c_2}$, where $\mathbb{1}$ is the monoidal unit (the trivial representation) in the category $\text{Rep}(k, G)$. The resulting tensor is obtained by evaluating the linear map $v$ at the identity $1 \in k$ and applying the transformation `fromPairT` to the resulting element of the tensor product $V_{c_1} \otimes_k V_{c_2}$. This construction defines "constant" or invariant rank-2 tensors, such as a metric or a unit tensor.

Let $S$ be a tensor species over a ring $k$ and a group $G$. For any index colors $c_1, c_2 \in C$, let $V_{c_1}$ and $V_{c_2}$ be the $k$-modules (representations) associated with these colors. Let $\mathbb{1}$ denote the monoidal unit (the trivial representation) in the category $\text{Rep}(k, G)$. For any morphism of representations $v: \mathbb{1} \to V_{c_1} \otimes_k V_{c_2}$ and any group element $g \in G$, the rank-2 tensor constructed via `fromConstPair` is invariant under the group action: $$g \cdot \text{fromConstPair}(v) = \text{fromConstPair}(v)$$

Let $S$ be a tensor species over a ring $k$ and a group $G$. For any colors $c_1, c_2, c_2' \in C$, let $V_{c_1}, V_{c_2}, V_{c_2'}$ be the $k$-modules (representations) associated with these colors. Let $h : c_2 = c_2'$ be an equality and $f : V_{c_2} \to V_{c_2'}$ be the $k$-linear map induced by $h$. Given a morphism $v : \mathbb{1} \to V_{c_1} \otimes_k V_{c_2}$ from the monoidal unit $\mathbb{1}$ in the category of representations $\text{Rep}(k, G)$, applying the rank-2 tensor construction `fromConstPair` to the composition of $v$ with the map $\text{id}_{V_{c_1}} \otimes f$ is equivalent to re-indexing the tensor constructed from $v$ using the identity permutation: $$\text{fromConstPair}(v \gg (\text{id}_{V_{c_1}} \otimes f)) = \text{permT}_{\text{id}}(\text{fromConstPair } v)$$ where $\text{permT}_{\text{id}}$ is the tensor permutation map corresponding to the identity permutation, which accounts for the change in index colors from configuration $[c_1, c_2]$ to $[c_1, c_2']$.

Let $S$ be a tensor species over a ring $k$ and a group $G$. For any two colors $c_1, c_2 \in C$, let $V_{c_1}$ and $V_{c_2}$ be the $k$-modules (representations) associated with these colors. Given a morphism $v: \mathbb{1} \to V_{c_1} \otimes_k V_{c_2}$ from the monoidal unit $\mathbb{1}$ in the category of representations $\text{Rep}(k, G)$, let $\beta_{V_{c_1}, V_{c_2}}: V_{c_1} \otimes V_{c_2} \to V_{c_2} \otimes V_{c_1}$ be the braiding (canonical commutativity isomorphism) in the category. The theorem states that applying the rank-2 tensor construction `fromConstPair` to the braided morphism $v \gg \beta$ is equivalent to taking the tensor constructed from $v$ and swapping its indices: $$\text{fromConstPair}(v \gg \beta) = \text{perm}_T(\sigma, h)(\text{fromConstPair}(v))$$ where $\sigma = [1, 0]$ is the permutation swapping the two indices and $h$ is the proof of consistency for the index colors.

For a tensor species $S$ over a ring $k$ and index colors $c_1, c_2, c_3 \in C$, `fromTripleT` is the $k$-linear map: $$\Phi : V_{c_1} \otimes_k (V_{c_2} \otimes_k V_{c_3}) \to S.\text{Tensor}(c_1, c_2, c_3)$$ where $V_{c_i} = (S.FD.obj(c_i)).V$ is the underlying $k$-module associated with the color $c_i$, and $S.\text{Tensor}(c_1, c_2, c_3)$ is the space of tensors with three indices of colors $c_1, c_2, c_3$. This map constructs a rank-3 tensor by mapping the triple tensor product of the component modules into the formal tensor space of the species.

Let $S$ be a tensor species over a ring $k$. For any index colors $c_1, c_2, c_3 \in C$ and elements $x \in V_{c_1}$, $y \in V_{c_2}$, and $z \in V_{c_3}$ (where $V_{c_i}$ is the module associated with color $c_i$), the rank-3 tensor constructed via $\text{fromTripleT}$ from the element $x \otimes (y \otimes z)$ is equal to the product of the rank-1 tensors $\text{fromSingleT } x$, $\text{fromSingleT } y$, and $\text{fromSingleT } z$ (as calculated by $\text{prodT}$), up to an identity permutation of the indices.

Let $S$ be a tensor species over a ring $k$ and a group $G$. For any index colors $c_1, c_2, c_3 \in C$, let $V_{c_i}$ be the $k$-module associated with the color $c_i$ and $\rho_{c_i}$ be the representation of $G$ on $V_{c_i}$. For any element $x \in V_{c_1} \otimes_k (V_{c_2} \otimes_k V_{c_3})$ and any group element $g \in G$, the action of $g$ on the rank-3 tensor constructed from $x$ via the linear map $\text{fromTripleT}$ is given by: $$g \cdot \text{fromTripleT}(x) = \text{fromTripleT}((\rho_{c_1}(g) \otimes (\rho_{c_2}(g) \otimes \rho_{c_3}(g)))x)$$ where $\cdot$ denotes the group action on the tensor space and $\otimes$ denotes the induced map on the tensor product of modules.

Let $S$ be a tensor species over a ring $k$ and $c, c_1, c_2 \in C$ be index colors. Let $V_c, V_{c_1}, V_{c_2}$ denote the $k$-modules associated with these colors. For any element $x \in V_c \otimes_k (V_{c_1} \otimes_k V_{c_2})$ and any multi-index $b$ for a rank-3 tensor with colors $(c, c_1, c_2)$, the component (coefficient) of the tensor $\text{fromTripleT}(x)$ in the tensor basis at index $b$ is equal to the component of $x$ in the product basis of $V_c \otimes (V_{c_1} \otimes V_{c_2})$ at indices $(b_0, b_1, b_2)$.