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Proof: “Otto” and “Otho” are orthogonal

Otto Deng

Claim

“Otto” and “Otho” are orthogonal with respect to a natural, symmetry-invariant inner product that encodes the central reflection of the 4-letter word while ignoring the boundary letters O,o.

Setup (vector space and encoding)

Let U be a real inner-product space with orthonormal basis \{e_t,e_h\}.

Interpret the interior letters t,h as e_t,e_h, and ignore the boundary O,o.

For any word of the form Oxyo with x,y\in\{t,h\}, encode it as

v_{Oxyo} \;=\; e_x\otimes e_y \;\in\; V := U\otimes U.

Thus

v_{\text{Otto}} = e_t\otimes e_t,\qquad v_{\text{Otho}} = e_t\otimes e_h.

Give V the standard tensor inner product

\langle u_1\!\otimes\!u_2,\; v_1\!\otimes\!v_2\rangle := \langle u_1,v_1\rangle_U\,\langle u_2,v_2\rangle_U,

extended bilinearly.

Group action (reflection)

Let G=\langle \sigma\rangle\cong C_2 act on V by swapping the two interior slots:

\sigma(u\otimes v) := v\otimes u.

This action is orthogonal (unitary) because the inner product on V is the tensor product of the orthonormal inner product on U .

Decomposition into irreducibles

The operator \sigma is an involutive, self-adjoint isometry on V with eigenvalues \pm1. Hence

V \;=\; V^{+}\;\oplus\; V^{-}, \quad V^{\pm} := \{ w\in V \mid \sigma w = \pm w\}.

These are the isotypic components for the two one-dimensional irreducible representations of C_2 (trivial and sign), and they are orthogonal:

\langle V^{+}, V^{-}\rangle = 0.

Concretely,

V^{+}=\text{Sym}^2(U),\qquad V^{-}=\wedge^2 U.

An explicit orthonormal basis is

\underbrace{e_t\!\otimes\!e_t,\;\frac{1}{\sqrt{2}}(e_t\!\otimes\!e_h+e_h\!\otimes\!e_t),\; e_h\!\otimes\!e_h}{\text{for }V^{+}} \quad\text{and}\quad \underbrace{\frac{1}{\sqrt{2}}(e_t\!\otimes\!e_h-e_h\!\otimes\!e_t)}{\text{for }V^{-}}.

Locate the two words inside the decomposition

  • v_{\text{Otto}}=e_t\otimes e_t\in V^{+}(it is fixed by \sigma).
  • v_{\text{Otho}}=e_t\otimes e_h decomposes as e_t\!\otimes\!e_h =\tfrac12\big(e_t\!\otimes\!e_h+e_h\!\otimes\!e_t\big) +\tfrac12\big(e_t\!\otimes\!e_h-e_h\!\otimes\!e_t\big) =: v_{\text{sym}} + v_{\text{alt}}, with v_{\text{sym}}\in V^{+} and v_{\text{alt}}\in V^{-}.

Orthogonality computation

By the orthogonal decomposition, \langle v_{\text{Otto}}, v_{\text{alt}}\rangle=0.

It remains to show \langle v_{\text{Otto}}, v_{\text{sym}}\rangle=0. Using the orthonormality \langle e_t,e_h\rangle_U=0,

\Big\langle e_t\!\otimes\!e_t,\; \tfrac12\big(e_t\!\otimes\!e_h+e_h\!\otimes\!e_t\big)\Big\rangle = \tfrac12\big(\langle e_t,e_t\rangle\langle e_t,e_h\rangle

• \langle e_t,e_h\rangle\langle e_t,e_t\rangle\big) = \tfrac12(1\cdot 0 + 0\cdot 1)=0.

  • Therefore, \langle v_{\text{Otto}}, v_{\text{Otho}}\rangle =\langle v_{\text{Otto}}, v_{\text{sym}}\rangle +\langle v_{\text{Otto}}, v_{\text{alt}}\rangle =0+0=0.

Conclusion

Under the natural, reflection-invariant inner product on the representation space V=U\otimes U of G\cong C_2, the encodings of “Otto” and “Otho” are orthogonal:

\boxed{\;\langle v_{\text{Otto}}, v_{\text{Otho}}\rangle = 0\;}

This uses precisely the representation-theoretic decomposition into the trivial and sign representations (with their orthogonality—an instance of Schur orthogonality) plus a direct check inside the symmetric component.

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