Notes on Tensor

18 Oct 2026 PDF

Basic Theorems.

1. $(V \otimes W)^* \simeq V^* \otimes W^*$
2. $\hom(V\otimes W, Z) \simeq \hom(V, \hom(W, Z))$
3. $V^{**} \simeq V$

Corollary. $V^* \otimes W \simeq \hom (V, W)$

Tensor

TFAE.

1. $M$ is a type $(n, m)$ tensor
2. $M \in V^{\otimes n} \otimes (V^*)^{\otimes m}$
3. $M: (V^*)^n \times V^m \to k$
4. $M: V^{\otimes m} \to V^{\otimes n}$

(1, 1)-tensor

Let $\Lambda$ be a (1, 1)-tensor, i.e. $\Lambda: V^* \times V \to k$.

If we fix a basis ${e_i}$ for $V$, we can write $\Lambda$ as a matrix form:

\[\Lambda^\mu_{\;\nu} =\Lambda(e^\mu, e_\nu)\]

For example, given:

\[\begin{gather} \Lambda(e^0, e_0) = a\\ \Lambda(e^1, e_0) = b\\ \Lambda(e^0, e_1) = c\\ \Lambda(e^1, e_1) = d, \end{gather}\]

we have:

\[\Lambda^\mu_{\;\nu} = \begin{pmatrix} a & c \\ b & d \end{pmatrix}\]

Note: Specification of matrix component is only relevant with the left-right order (rightmost index corresponds to column) and irrelevant with the up-down position.

Let $\mathbf{a} = a^\mu e_\mu \in V$. Since $\Lambda(-, \mathbf{a}) : V^* \to k$ and $\hom(V^*, k) \simeq V$, we have $\Lambda(-, \mathbf{a}) \in V$. Write $\Lambda(-, \mathbf{a}) = \mathbf{a}’ = (a’)^\mu e_\mu$. We then have:

\[(a')^\mu = \Lambda(e^\mu, \mathbf{a}) = \sum_\nu \Lambda(e^\mu, e_\nu)a^\nu = \Lambda^\mu_{\;\nu}a^\nu.\]

(0, 2)-tensor

Let $\eta$ be a (0, 2)-tensor, i.e. $\eta: V \times V \to k$.

If we fix a basis ${e_i}$ for $V$, we can write $\eta$ as a matrix form:

\[\eta_{\mu \nu} =\eta(e_\mu, e_\nu)\]

For example, given:

\[\begin{gather} \eta(e_0, e_0) = a\\ \eta(e_1, e_0) = b\\ \eta(e_0, e_1) = c\\ \eta(e_1, e_1) = d, \end{gather}\]

we have:

\[\eta_{\mu\nu} = \begin{pmatrix} a & c \\ b & d \end{pmatrix}\]

Let $\mathbf{a} = a^\mu e_\mu \in V$. Since $\eta(-, \mathbf{a}) : V \to k$ and $\hom(V, k) \simeq V^*$, we have $\Lambda(-, \mathbf{a}) \in V^*$. Write $\Lambda(-, \mathbf{a}) = \mathbf{a}’ = (a’)_\mu e^\mu$. We then have:

\[(a')_\mu = \eta(e_\mu, \mathbf{a}) = \sum_\nu \eta(e_\mu, e_\nu)a^\nu = \eta_{\mu\nu}a^\nu.\]

Generalizing, for a $(2, 3)$-tensor $T^{\alpha \beta}_{\gamma \delta \epsilon} : V^* \times V^* \times V \times V \times V \to k$, if we product $T$ with 1 dual coordinate $a_\mu$ and 3 vector coordinates $b^\nu$, then we are left with a coordinate for $\hom(V^*, k) \simeq V$, which is a contravariant result. Indeed,

\[T^{\alpha \beta}_{\gamma \delta \epsilon} b_\beta c^\gamma d^\delta e^\epsilon = a^\alpha\]

Alternatively we can product $T$ with 2 dual coordinates and 2 vector coordinates to obtain a coordinate for $\hom(V, k) \simeq V^*$.

Index raising/lowering

Let $M^\mu_{\;\nu}$ act on $V$. How does $M$ “act” on $V^*$?

1. $\phi(v)$ conservation

If we require that $\phi(v)$ be conserved, i.e.

\[(M^*\phi)(Mv) = \phi(v),\]

then $M$ should act inversely on $V^*$. That is,

\[\begin{gather} (M^*)_\mu^{\;\nu} a_\nu = a'_\mu \\\\ \text{where} \quad M^* := (M^{-1})^T \end{gather}\]

2. Coordinate change

Alternatively, we can simply change $\phi$ to its corresponding vector $v_\phi$, transform $v_\phi$, and change it back to $V^*$. That is,

\[\begin{gather} (M')_\mu^{\;\nu}a_\nu := \eta_{\mu \sigma}( M^\sigma_{\;\tau} a^\tau) = \eta_{\mu\sigma} M^\sigma_{\;\tau}\eta^{\tau\nu}a_\nu \\\\ \text{i.e.} \quad M' = \eta M \eta \end{gather}\]

Theorem. If $\eta$ is invariant under $M$, i.e. $\eta(Mu, Mv) = \eta(u, v)$, the two conversions coincide.

Hence, we work with 2, since it’s much easier to calculate. The point is that, insofar we deal with Lorentz invariant transformations, index raising/lowering and contraction of indices flow naturally.

Furthermore, since $\eta$ has nonzero indices only on the diagonal, we can use the following shortcut:

Multiply by -1 when raising/lowering time index. Leave everything else unchanged.

Example.

\[\begin{gather} \Lambda^\mu_{\;\nu} = \begin{pmatrix} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \\\\ \Lambda_{\mu\nu} = \begin{pmatrix} -\gamma & \beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \\\\ \Lambda^{\mu\nu} = \begin{pmatrix} -\gamma & -\beta\gamma & 0 & 0\\ \beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\\\\ \Lambda_{\mu}^{\;\nu} = \begin{pmatrix} \gamma & \beta\gamma & 0 & 0\\ \beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \end{gather}\]